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class="expanded parent selected"><label style="padding-left: 14px"><i class="fa fa-fw"></i><a href="capacitors.html">Capacitors</a></label></li></ul></li></ul></div> <div class="hidden-xs hidden-sm"> <button class="btn btn-default btn-block text-xs-center" data-toggle="modal" data-target="#modal-feedback" style="margin-bottom: 10px"><i class="fa fa-send"></i>&nbsp;&nbsp;Feedback</button> </div> </div> <div class="col-md-9" id="main-column"> <h1 class="page_title"> Capacitors <a href="#" class="mark-page-favorite pull-right" data-pid="1136" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../physics.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../445/electricity-and-magnetism.html">Electricity and magnetism</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../1144/ahl-em-induction.html">AHL EM induction</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Capacitors</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 50 minutes"><i class="fa fa-clock-o"></i> 50&apos;</span> </ol> <article id="main-article"> <p><img alt="" src="../../em-induction/capacitor.jpg" style="float: left; width: 250px; height: 150px;">Capacitors are electrical components used in circuits requiring timing or smoothing. They store charge, which is then released at a rate that can be controlled.</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <div> <p>A capacitor consists of two parallel plates separated by a dielectric material. Current cannot flow through the dielectric; instead, charge is stored on the plates. The circuit symbol of a capacitor is effective!</p> <p style="text-align: center;"><img alt="" src="../../em-induction/cap-sym.jpg" style="width: 100px; height: 61px;"></p> <p>The capacitance of a capacitor is defined as the ratio of the charge stored to the potential difference across the capacitor:</p> <p style="text-align: center;"><span class="math-tex">\(C={q\over V}\)</span></p> <ul> <li><span class="math-tex">\(C\)</span>&nbsp;is capacitance (F)</li> <li><span class="math-tex">\(q\)</span>&nbsp;is charge (C)</li> <li><span class="math-tex">\(V\)</span>&nbsp;is potential difference (V)</li> </ul> <p>For a fixed capacitance,&nbsp;<span class="math-tex">\(q=CV\)</span>. Capacitance can be determined from the gradient of a charge-potential difference graph.</p> <p style="text-align: center;"><img alt="" src="../../em-induction/q-v.png" style="width: 250px; height: 218px;"></p> <div class="panel panel-turquoise panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Work done and charge</p> </div> </div> <div class="panel-body"> <div> <p>The work required to add charges to already charged plates increases with the charge stored.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369634398"></iframe></div> <p>It is not only parallel plates that can store charges. A sphere can also store charge.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369634334"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p>&nbsp;</p> </div> </div> </div> <div class="panel panel-turquoise panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Energy stored</p> </div> </div> <div class="panel-body"> <div> <p>The energy stored when a capacitor is charged can be calculated by adding the infinitesimal stages of work done when each new quantity of charge is added. This is equal to the area under the graph.&nbsp;Since charge stored is proportional to the potential difference across the capacitor:</p> <p style="text-align: center;"><span class="math-tex">\(E={1\over 2}qV\)</span></p> <p style="text-align: center;"><span class="math-tex">\(\Rightarrow E={1\over 2}CV^2\)</span></p> <ul> <li><span class="math-tex">\(E\)</span>&nbsp;is the energy stored on the capacitor or the work done that was required to store the charges (J)</li> </ul> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369633619"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p>&nbsp;</p> </div> </div> </div> <div class="panel panel-turquoise panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Charging and discharging</p> </div> </div> <div class="panel-body"> <div> <h4>Circuit diagram</h4> <p>Capacitors are charged when connected in series with a power supply (position A in the diagram below). The charge stored is at a maximum when the potential difference on the capacitor rises to match the power supply. At this point, no current flows.</p> <p>Capacitors discharge through a given resistor when not connected to a power supply (position B). The current is maximum at first. As the charge stored on the capacitor decreases, so does the potential difference across it. The current falls.</p> <p style="text-align: center;"><img alt="" src="../../em-induction/discharge-capacitor-circuit-diagram-1.png" style="width: 250px; height: 264px;"></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369633328"></iframe></div> <h4>Exponential relationships</h4> <p>The repelling nature of the charges stored has the effect that the rate of discharge from a capacitor is proportional to the charge remaining on a capacitor. Mathematically,&nbsp;<span class="math-tex">\(-{\mathrm{d}q\over \mathrm{d}t}\propto q\)</span>. Expressions for the variation of charge, current and potential difference with time can be derived:</p> <p style="text-align: center;"><span class="math-tex">\(q=q_0e^{-t\over \tau}\)</span></p> <p style="text-align: center;"><span class="math-tex">\(I=I_0e^{-t\over \tau}\)</span></p> <p style="text-align: center;"><span class="math-tex">\(V=V_0e^{-t\over \tau}\)</span></p> <ul> <li><span class="math-tex">\(q\)</span>&nbsp;is the remaining charge stored on the capacitor (C)</li> <li><span class="math-tex">\(q_0\)</span>&nbsp;is the initial charge stored on the capacitor (C)</li> <li><span class="math-tex">\(I\)</span>&nbsp;is the current&nbsp;(A)</li> <li><span class="math-tex">\(I_0\)</span>&nbsp;is the initial current&nbsp;(A)</li> <li><span class="math-tex">\(V\)</span>&nbsp;is the potential difference across the capacitor (C)</li> <li><span class="math-tex">\(V_0\)</span>&nbsp;is the initial potential difference across the capacitor (C)</li> <li><span class="math-tex">\(t\)</span>&nbsp;is the time over which the capacitor has been discharging (s)</li> <li><span class="math-tex">\(\tau\)</span>&nbsp;is the time constant (s)</li> </ul> <p>These equations are exponential decay relationships and have the following characteristic shape when plotted against time:</p> <p style="text-align: center;"><img alt="" src="../../em-induction/discharge.png" style="width: 250px; height: 141px;"></p> <p>When a capacitor is charged:</p> <ul> <li>the charge stored and potential difference increase (blue)</li> <li>current decreases (green)</li> </ul> <p style="text-align: center;"><img alt="" src="../../em-induction/charging.png" style="width: 250px; height: 131px;"></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369633407"></iframe></div> <h4>Time constant</h4> <p>The time constant for a capacitor circuit is defined as the time taken for the charge, current or potential difference to fall to&nbsp;<span class="math-tex">\(1\over e\)</span>&nbsp;of their original value. This is approximately&nbsp;<span class="math-tex">\(0.37q_0\)</span>,&nbsp;<span class="math-tex">\(0.37I_0\)</span>&nbsp;or&nbsp;<span class="math-tex">\(0.37 V_0\)</span>.</p> <p>The time constant can be found from the graph in one of two ways:</p> <ul> <li>Moving down the vertical axis to find&nbsp;<span class="math-tex">\(0.37q_0\)</span>, etc, and then going across to the curve and down to the time.</li> <li>Drawing a tangent to the curve at&nbsp;<span class="math-tex">\(t=0\)</span>&nbsp;and extrapolating until it cuts the time axis.</li> </ul> <p>Time constant also has a real-world meaning. It is the product of the capacitance and load resistance (for the circuit diagram shown above):</p> <p style="text-align: center;"><span class="math-tex">\(\tau=RC\)</span></p> <ul> <li><span class="math-tex">\(\tau\)</span>&nbsp;is time constant (s)</li> <li><span class="math-tex">\(R\)</span>&nbsp;is load resistance (<span class="math-tex">\(\Omega\)</span>)</li> <li><span class="math-tex">\(C\)</span>&nbsp;is capacitance (F)</li> </ul> </div> </div> </div> <div class="panel panel-turquoise panel-has-colored-body panel-has-border panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Combinations</p> </div> </div> <div class="panel-body"> <div> <p>Like other circuit components, capacitors can be connected in series and in parallel.</p> <p>In parallel, the potential difference on each capacitor is in the same direction; each stores charge in the same orientation. The total capacitance for capacitors in parallel is the sum of the individual capacitances:</p> <p style="text-align: center;"><span class="math-tex">\(C_\text{parallel}=C_1+C_2+...+C_n\)</span></p> <p style="text-align: center;"><img alt="" src="../../em-induction/parallel.png" style="width: 250px; height: 147px;"></p> <p>In series, the capacitors counteract one another. The total capacitance is less than any individual capacitor:</p> <p style="text-align: center;"><span class="math-tex">\({1\over C_\text{series}}={1\over C_1}+{1\over C_2}+...+{1\over C_n}\)</span></p> <p style="text-align: center;"><img alt="" src="../../em-induction/series.png" style="width: 250px; height: 68px;"></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369633259"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p>&nbsp;</p> </div> </div> </div> </div> </div> <div class="panel-footer"> <div>&nbsp;</div> </div> </div> <div class="panel panel-has-colored-body panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <p>A capacitor contains a uniform electric field. The value of the capacitance of a capacitor is set according to three variables:</p> <p style="text-align: center;"><span class="math-tex">\(C=\varepsilon{A\over d}\)</span></p> <ul> <li><span class="math-tex">\(C\)</span>&nbsp;is capacitance (F)</li> <li><span class="math-tex">\(\varepsilon\)</span>&nbsp;is an electric constant (Fm<sup>&minus;1</sup>)</li> <li><span class="math-tex">\(A\)</span> is the area of overlap of the two plates (m<sup>2</sup>)</li> <li><span class="math-tex">\(d\)</span> is the separation between the plates (m)</li> </ul> <p>The electric constant is the product of the permittivity of free space and the relative permittivity of the dielectric:</p> <p style="text-align: center;"><span class="math-tex">\(\varepsilon=\varepsilon_0k\)</span></p> <ul> <li><span class="math-tex">\(\varepsilon_0\)</span>&nbsp;is the permittivity of space (8.854 x 10<sup>-12</sup> Fm<sup>-1</sup>)</li> <li><span class="math-tex">\(k\)</span>&nbsp;is the permittivity of the dielectric (~1 for air and &gt;1 for all media)</li> </ul> <div class="panel panel-has-colored-body panel-has-border panel-expandable panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Area and distance between the plates</p> </div> </div> <div class="panel-body"> <div> <p>The capacitance of a capacitor can be increased by:</p> <ul> <li>increasing the area of the plates (often achieved using a spiral shape)</li> <li>decreasing the separation of the plates</li> </ul> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369634556"></iframe></div> <p>A common exam question will alter the plate separation once they have already been charged. Work is done to separate the attracting plates.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369634866"></iframe></div> <p>Another common question involves charging or discharging the capacitor with constant current - watch out! These are easy to deal with. Simply revert to the usual equations, such as&nbsp;<span class="math-tex">\(I={q\over t}\)</span>.</p> </div> </div> <div class="panel-footer"> <div> <p>&nbsp;</p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-expandable panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Effect of the dielectric</p> </div> </div> <div class="panel-body"> <div> <p>The nature of the dielectric material also affects the capacitance.&nbsp;</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/369634474"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p>&nbsp;</p> </div> </div> </div> <p>&nbsp;</p> </div> <div class="panel-footer"> <div>&nbsp;</div> </div> </div> <div class="panel panel-has-colored-body panel-purple"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Summary</p> </div> </div> <div class="panel-body"> <div> <div class="panel panel-has-colored-body panel-has-border panel-purple"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Rectification</p> </div> </div> <div class="panel-body"> <div> <p>Capacitors can be used in rectification to overcome the sinusoidal limitation of full wave diode rectification. The presence of a capacitor in parallel with the load smooths the current through the resistor by continuing to release charge even when the EMF across the power supply falls to zero. This has a smoothing effect.</p> <p style="text-align: center;"><img alt="" src="../../em-induction/bridge.png" style="width: 500px; height: 190px;"></p> <p>This bridge circuit can appear to be complex at first. However,&nbsp;tracing the flow of current upward from the power supply through A, D, R, C, B and back to the bottom of the power supply shows that diode bias is adhered to throughout.</p> </div> </div> <div class="panel-footer"> <div> <p>&nbsp;</p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-purple panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Wien bridge circuit</p> </div> </div> <div class="panel-body"> <div> <p>Just as a Wheatstone bridge circuit enabled determination of an unknown resistance, a Wien bridge circuit can be used to determine an unknown capacitance.</p> <p style="text-align: center;"><img alt="" src="../../em-induction/wien.svg" style="width: 250px; height: 173px;"></p> </div> </div> <div class="panel-footer"> <div> <p>&nbsp;</p> </div> </div> </div> </div> </div> <div class="panel-footer"> <div>&nbsp;</div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <p><em>Use quizzes&nbsp;to practise application of theory.</em></p> <div class="panel panel-has-colored-body panel-has-border panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Capacitance</p> </div> </div> <div class="panel-body"> <div> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#a216274a"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="a216274a"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Capacitance <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="6-353-1136" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>The image shows two metal spheres.</p><p style="text-align: center;"><img alt="" height="144" src="../../screenshot-2019-09-30-at-06.08.09.png" width="271"></p><p>Charge +<span class="math-tex">\(Q\)</span> is added to the large sphere. What charge should be added to the small sphere so that it is at the same potential?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(4Q\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(Q\over4\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(Q\over2\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(2Q\)</span></span></label> </p></div><div class="q-explanation"><p>Electrical potential, <span class="math-tex">\(V = {Q\over 4πε_0r}\)</span></p><p>For the same potential, <span class="math-tex">\(Q\propto r\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The image shows two metal spheres.</p><p style="text-align: center;"><img alt="" height="144" src="../../screenshot-2019-09-30-at-06.08.09.png" width="271"></p><p>The capacitance of the large sphere is 10 pF. What is the capacitance of the small sphere?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>√10 pF</span></label> </p><p><label class="radio"> <input type="radio"> <span>20 pF</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 pF</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>5 pF</span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(C = 4πε_0r\Rightarrow C\propto r\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The image shows two metal spheres.</p><p style="text-align: center;"><img alt="" height="144" src="../../screenshot-2019-09-30-at-06.08.09.png" width="271"></p><p>Adding charge <span class="math-tex">\(+Q\)</span> to the large sphere raises its potential to 10 V. What will the potential difference between the spheres be if <span class="math-tex">\(-Q\)</span> is added to the small sphere?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>15 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>20 V</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>30 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>10 V</span></label> </p></div><div class="q-explanation"><p>The small one has half the radius and so half the capacitance. For the same charge, <span class="math-tex">\(V\propto {1\over r}\)</span> and so the potential is -20 V.</p><p>The difference in potential <span class="math-tex">\(= 10-(-20) = 30 \text{ V}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>An adjustable circular parallel plate capacitor has capacitance 16 pF.</p><p>What would the capacitance be if both plate separation and plate radius are doubled?</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>32 pF</span></label> </p><p><label class="radio"> <input type="radio"> <span>8 pF</span></label> </p><p><label class="radio"> <input type="radio"> <span>16 pF</span></label> </p><p><label class="radio"> <input type="radio"> <span>64 pF</span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(C\propto {A\over d}\propto {r^2\over d}\)</span></p><p>The overall effect is x2.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>An adjustable circular parallel plate capacitor has capacitance 16 pF.</p><p>A dielectric is put between the plates and their separation is reduced to one fifth of the original. If the new capacitance is 0.16 nF, what is the dielectric constant of the material?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>5 Fm<sup>-1</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>20 Fm<sup>-1</sup></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>2 Fm<sup>-1</sup></span></label> </p><p><label class="radio"> <input type="radio"> <span>50 Fm<sup>-1</sup><span class="scayt-misspell-word" data-scayt-word="answer5" data-wsc-lang="en_US"></span></span></label> </p></div><div class="q-explanation"><p>Since area is constant, <span class="math-tex">\(C \propto {ε\over d }\)</span></p><p>The capacitance is 10x bigger, so <span class="math-tex">\(ε\over d\)</span> must be 10x bigger. The reduction in separation contributes 5x and so the dielectric constant must be 2x the dielectric constant of air (1 Fm<sup>-1</sup>).</p><p>NB: Don't forget to check unit prefixes p = 10<sup>-12</sup> n = 10<sup>-9</sup></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Charge <span class="math-tex">\(Q\)</span> is transferred to the plates of an adjustable parallel plate capacitor by connecting to a battery of EMF <span class="math-tex">\(V\)</span>. If the plate separation is doubled while still connected to the battery, what will the potential difference across and charge stored on the plates be?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(2V\text{, }{Q}\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({V\over 2}\text{, }{Q\over 2}\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(V\text{, }{Q\over 2}\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(V\text{, }{2Q}\)</span></span></label> </p></div><div class="q-explanation"><p>Since the capacitor is connected to the same battery, <span class="math-tex">\(V\)</span> is the same.</p><p>The capacitance has halved (<span class="math-tex">\(C\propto {1\over d}\)</span>), the charge stored will half (<span class="math-tex">\(q\propto C\)</span>​​​​​​​).</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Charge <span class="math-tex">\(Q \)</span> is transferred to the plates of an adjustable parallel plate capacitor by connecting to a battery of EMF <span class="math-tex">\(V\)</span>. If the battery is disconnected and the plate separation doubled, what will the potential difference across and charge stored on the plates be?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({V\over2}\text{, }Q\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(V\text{, }2Q\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(2V\text{, }Q\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(2V\text{, } 2Q\)</span></span></label> </p></div><div class="q-explanation"><p>The plates are isolated so <span class="math-tex">\(Q\)</span> is constant so field is constant.</p><p>The distance between plates is doubled, so the work done moving between them has been doubled. This implies that potential difference is doubled.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The effect of placing a dielectric between the plates of a capacitor is to reduce the:</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>potential difference across the plates for a given charge</span></label> </p><p><label class="radio"> <input type="radio"> <span>permittivity of the medium between the plates</span></label> </p><p><label class="radio"> <input type="radio"> <span>charge for a given potential difference</span></label> </p><p><label class="radio"> <input type="radio"> <span>capacitance</span></label> </p></div><div class="q-explanation"><p>The placement of the dielectric implies that the previous material was air, with a permittivity of 1. This can only increase the permittivity and, therefore, the capacitance since <span class="math-tex">\(C\propto \varepsilon\)</span></p><p><span class="math-tex">\(Q=CV\)</span> so charge would be increased for a given potential difference</p><p><span class="math-tex">\(V={Q\over C}\)</span> so potential difference would be reduced for a given charge</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two capacitors are connected in series.</p><p style="text-align: center;"><img alt="" height="149" src="../../screenshot-2019-09-30-at-11.31.25.png" width="148"></p><p>Calculate the total capacitance.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>0.25 pF</span></label> </p><p><label class="radio"> <input type="radio"> <span>18 pF</span></label> </p><p><label class="radio"> <input type="radio"> <span>6 pF</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>4 pF</span></label> </p></div><div class="q-explanation"><p>Try this without a calculator!</p><p><span class="math-tex">\({1\over C} = {1\over C_1} + {1\over C_2} = {1\over 12} + {1\over 6} = {1\over12} + {2\over12} = {3\over 12}={1\over 4}\)</span></p><p><span class="math-tex">\(C_\text{series} = 4\text{ pF}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two capacitors are connected in series.</p><p style="text-align: center;"><img alt="" height="149" src="../../screenshot-2019-09-30-at-11.31.25.png" width="148"></p><p>Calculate the total charge stored.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>6 pC</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>24 pC</span></label> </p><p><label class="radio"> <input type="radio"> <span>4 pC</span></label> </p><p><label class="radio"> <input type="radio"> <span>12 pC</span></label> </p></div><div class="q-explanation"><p>​​​​​​​<span class="math-tex">\({1\over C} = {1\over C_1} + {1\over C_2} = {1\over 12} + {1\over 6} = {1\over12} + {2\over12} = {3\over 12}={1\over 4}\)</span></p><p><span class="math-tex">\(C_\text{series} = 4\text{ pF}\)</span></p><p><span class="math-tex">\(Q = CV = 4 \times 6=24\text{ C}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two capacitors are connected in series.</p><p style="text-align: center;"><img alt="" height="149" src="../../screenshot-2019-09-30-at-11.31.25.png" width="148"></p><p>Calculate the charge stored on the 12 pF capacitor.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>4 pC</span></label> </p><p><label class="radio"> <input type="radio"> <span>6 pC</span></label> </p><p><label class="radio"> <input type="radio"> <span>12 pC</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>24 pC</span></label> </p></div><div class="q-explanation"><p>No charge is stored in the gap between the capacitors since it cannot be added or taken away from the conductor. Therefore the charge is the same as it is for the combination.</p><p>​​​​​​​<span class="math-tex">\({1\over C} = {1\over C_1} + {1\over C_2} = {1\over 12} + {1\over 6} = {1\over12} + {2\over12} = {3\over 12}={1\over 4}\)</span></p><p><span class="math-tex">\(C_\text{series} = 4\text{ pF}\)</span></p><p><span class="math-tex">\(Q = CV = 4 \times 6=24\text{ C}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two capacitors are connected in series.</p><p style="text-align: center;"><img alt="" height="149" src="../../screenshot-2019-09-30-at-11.31.25.png" width="148"></p><p>Calculate the potential difference across the 12 pF capacitor.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>6 V</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>2 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>4 V</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.5 V</span></label> </p></div><div class="q-explanation"><p>To start we will find the charge on the 12 pF capacitor, which is the same as the charge stored for the combination.</p><p>​​​​​​​<span class="math-tex">\({1\over C} = {1\over C_1} + {1\over C_2} = {1\over 12} + {1\over 6} = {1\over12} + {2\over12} = {3\over 12}={1\over 4}\)</span></p><p><span class="math-tex">\(C_\text{series} = 4\text{ pF}\)</span></p><p><span class="math-tex">\(Q = CV = 4 \times 6=24\text{ C}\)</span></p><p>And now the potential difference <span class="math-tex">\(V = {Q\over C} = {24\over 12} = 2\text{ V}\)</span>​​​​​​​</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two capacitors are connected in parallel.</p><p style="text-align: center;"><img alt="" height="195" src="../../screenshot-2019-09-30-at-11.55.00.png" width="127"></p><p>Calculate the total capacitance.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>4 pF</span></label> </p><p><label class="radio"> <input type="radio"> <span>6 pF</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>18 pF</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.25 pF</span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(C = C_1 + C_2\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two capacitors are connected in parallel.</p><p style="text-align: center;"><img alt="" height="195" src="../../screenshot-2019-09-30-at-11.55.00.png" width="127"></p><p>Calculate the charge on the 6 pF capacitor.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>18 pC</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>36 pC</span></label> </p><p><label class="radio"> <input type="radio"> <span>1 pC</span></label> </p><p><label class="radio"> <input type="radio"> <span>324 pC</span></label> </p></div><div class="q-explanation"><p>The potential difference acts across each of the capacitors.</p><p><span class="math-tex">\(Q = CV=6\text{ pF}\times 6\text{ V}=36\text{ pC}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two capacitors are connected in parallel.</p><p style="text-align: center;"><img alt="" height="195" src="../../screenshot-2019-09-30-at-11.55.00.png" width="127"></p><p>Calculate the ratio <span class="math-tex">\(V_\text{12 pF} \over V_\text{6 pF}\)</span>.</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(2\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(4\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(1\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(1\over2\)</span></span></label> </p></div><div class="q-explanation"><p>The potential difference across each of the capacitors is the same because each is in parallel with the power supply.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> <i class="fa fa-arrow-left"></i>&nbsp;&nbsp;Prev </button> </div> <div class="pull-right pull-xs-none"> <button class="btn btn-success btn-xs-block text-xs-center btn-results" style="display: none"> <i class="fa fa-bar-chart"></i> Check Results </button> <button class="btn btn-default d-xs-none btn-next"> Next&nbsp;&nbsp;<i class="fa fa-arrow-right"></i> </button> <button class="btn btn-default btn-xs-block text-xs-center btn-close" data-dismiss="modal" style="display: none"> Close </button> </div> </div> </div> </div> </div></div> </div> </div> <div class="panel-footer"> <div> <p>&nbsp;</p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Charging and discharging capacitors</p> </div> </div> <div class="panel-body"> <div> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#ea4c9fc3"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="ea4c9fc3"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Charging and discharging capacitors <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="6-354-1136" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>The image is from the PhET circuit simulator. The right hand side shows a list of possible potential difference graphs for each of the voltmeters connected in the circuit.</p><p style="text-align: center;"><img alt="" height="325" src="../../screenshot-2019-10-02-at-15.05.01.png" width="393"></p><p>The capacitor is uncharged. Vb is connected across a cell.</p><p>Fill in the letters representing the graphs of the changing potential differences when switch 2 is closed.</p></div><div class="q-answer"><p>V1 <input type="text" style="height: auto;" data-c="A"> <span class="review"></span></p><p>Vr <input type="text" style="height: auto;" data-c="B"> <span class="review"></span></p><p>Vc <input type="text" style="height: auto;" data-c="E"> <span class="review"></span></p><p>V2 <input type="text" style="height: auto;" data-c="C"> <span class="review"></span></p><p>Vb <input type="text" style="height: auto;" data-c="D"> <span class="review"></span></p></div><div class="q-explanation"><p>Watch and see! Notice the shape of the rising potential difference across the capacitor and the corresponding decrease for the resistor in series.</p><p style="text-align: center;"><img alt="" height="342" class="gifffer" data-gifffer="/media/physics/capchq1.gif" width="341"></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The image is from the PhET circuit simulator. The right hand side shows a list of possible potential difference graphs for each of the voltmeters connected in the circuit.</p><p style="text-align: center;"><img alt="" height="283" src="../../screenshot-2019-10-02-at-16.05.01.png" width="347"></p><p>The capacitor is charged.</p><p>Fill in the letters representing the graphs of the changing potential differences when switch 1 (the top switch) is closed.</p></div><div class="q-answer"><p>V1 <input type="text" style="height: auto;" data-c="C"> <span class="review"></span></p><p>Vr <input type="text" style="height: auto;" data-c="A"> <span class="review"></span></p><p>Vc <input type="text" style="height: auto;" data-c="D"> <span class="review"></span></p><p>V2 <input type="text" style="height: auto;" data-c="B"> <span class="review"></span></p><p>Vb <input type="text" style="height: auto;" data-c="E"> <span class="review"></span></p></div><div class="q-explanation"><p>Watch and see! Notice the shape of the decreasing potential difference across the capacitor and the corresponding increase for the resistor in series.</p><p style="text-align: center;"><img alt="" height="309" class="gifffer" data-gifffer="/media/physics/discq2.gif" width="309"></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A capacitor is charged through a resistor. The final potential difference across the plates is <span class="math-tex">\(V\)</span> and the time taken to reach <span class="math-tex">\({2\over3}V\)</span> is <span class="math-tex">\(T\)</span> seconds.</p><p style="text-align: center;"><img alt="" height="176" src="../../screenshot-2019-10-02-at-17.05.59.png" width="198"></p><p>The separation of the plates is doubled. The final potential difference and the time to reach <span class="math-tex">\(2\over3\)</span> of that value is now:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(V\text{, }2T\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(V\text{, }{T\over2}\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(2V\text{, }2T\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\({V\over2}\text{, }{T\over2}\)</span></span></label> </p></div><div class="q-explanation"><p>The capacitance is halved since <span class="math-tex">\(C\propto{1\over d}\)</span>, so <span class="math-tex">\(CR\)</span> is halved. This means the time constant (approximately the time needed for a charging capacitor's potential difference to rise to <span class="math-tex">\(2\over3\)</span> of its final value) halves.</p><p>The final potential difference is unchanged as the power supply is unchanged.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>A capacitor is charged through a resistor.</p><p style="text-align: center;"><img alt="" height="176" src="../../screenshot-2019-10-02-at-17.05.59.png" width="198"></p><p>If <span class="math-tex">\(C\)</span> is 10 μF and it takes 20 s to reach 2/3 of its final charge, what is resistance <span class="math-tex">\(R\)</span>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>20 MΩ</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>2 MΩ</span></label> </p><p><label class="radio"> <input type="radio"> <span>0.2 MΩ</span></label> </p><p><label class="radio"> <input type="radio"> <span>20 kΩ</span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(\tau=CR = 20\text{ s}\)</span></p><p><span class="math-tex">\(R = {20\over10\times10^{-6}} = 2 \times 10^6\text{ Ω}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two capacitors are connected in series.</p><p style="text-align: center;"><img alt="" height="215" src="../../screenshot-2019-09-30-at-11.31.25.png" width="213"></p><p>What is the ratio <span class="math-tex">\(\text{energy stored}_\text{12 pF} \over\text{energy stored}_\text{6 pF}\)</span>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(1\over4\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(4\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(2\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(1\over2\)</span></span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(E = {1\over2}{Q^2\over C}\)</span> and <span class="math-tex">\(Q\)</span> is the same for each</p><p><span class="math-tex">\(\Rightarrow E\propto{1\over C}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>Two capacitors are connected in parallel.</p><p style="text-align: center;"><img alt="" height="212" src="../../screenshot-2019-09-30-at-11.55.00.png" width="137"></p><p>What is the ratio <span class="math-tex">\(\text{energy stored}_\text{12 pF} \over\text{energy stored}_\text{6 pF}\)</span>?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(1\over2\)</span></span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(2\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(4\)</span></span></label> </p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(1\over4\)</span></span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(E = {1\over 2}CV^2\)</span> and <span class="math-tex">\(V\)</span> is the same for each</p><p><span class="math-tex">\(\Rightarrow E\propto C\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>The graph shows the potential difference across a 2 μF capacitor as it discharges through a resistor, R.</p><p style="text-align: center;"><img alt="" height="168" src="../../electricity/screenshot-2019-10-03-at-06.11.10.png" width="185"></p><p>What is the resistance of R?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>0.5 MΩ</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>2 MΩ</span></label> </p><p><label class="radio"> <input type="radio"> <span>1.25 MΩ</span></label> </p><p><label class="radio"> <input type="radio"> <span>20 MΩ</span></label> </p></div><div class="q-explanation"><p>The time taken to reach <span class="math-tex">\(1\over3\)</span> of initial potential difference, <span class="math-tex">\(\tau =RC = 4\)</span></p><p><span class="math-tex">\(R={4\over 2\times10^{-6}}=2\times 10^6\text{ }\Omega\)</span>​​​​​​​</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> <i class="fa fa-arrow-left"></i>&nbsp;&nbsp;Prev </button> </div> <div class="pull-right pull-xs-none"> <button class="btn btn-success btn-xs-block text-xs-center btn-results" style="display: none"> <i class="fa fa-bar-chart"></i> Check Results </button> <button class="btn btn-default d-xs-none btn-next"> Next&nbsp;&nbsp;<i class="fa fa-arrow-right"></i> </button> <button class="btn btn-default btn-xs-block text-xs-center btn-close" data-dismiss="modal" style="display: none"> Close </button> </div> </div> </div> </div> </div></div> </div> </div> <div class="panel-footer"> <div> 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