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<p> In this lesson we explore our first circuit element - capacitors.</p><p>Note that in most of the videos in this lesson, the potential is written as \(\phi\) instead of \(V\).</p><p> Textbook Links </p><ul><li><a href="/courses/course-v1:MITx+8.02.1x+1T2019/pdfbook/0/chapter/5/3">Chapter 5.1-5.3 Capacitors</a></li></ul>
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<h2 class="hd hd-2 unit-title">L12Q1: Capacitance of the Earth</h2>
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Capacitance of the Earth
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What is the Capacitance of the Earth? Assume the earth is a spherical conductor with a radius of [mathjaxinline]6.4\times 10^{6}[/mathjaxinline]m and use [mathjaxinline]k\equiv \dfrac {1}{4\pi \varepsilon _0}=9\times 10^9\, N \cdot m^2 / C^2[/mathjaxinline]. Express your answer in [mathjaxinline]mF[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]C=[/mathjaxinline]</p>
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Potential Difference
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Consider two uniformly charged parallel plates with opposite polarity as shown. The electric potential difference [mathjaxinline]V(B)-V(A)[/mathjaxinline] between the points [mathjaxinline]A[/mathjaxinline] and [mathjaxinline]B[/mathjaxinline] that lie outside the plates is </p>
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<h2 class="hd hd-2 unit-title">L12Q3: Properties of a Parallel Plate Capacitor</h2>
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Properties of a Parallel Plate Capacitor
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<p><b> (Part a) </b> Suppose you wish to construct a [mathjaxinline]1\, F [/mathjaxinline] capacitor from two square parallel plates that are a
distance of [mathjaxinline]d=1\,cm [/mathjaxinline] apart. What side
length [mathjaxinline]L[/mathjaxinline] (in meters) is required? Use [mathjaxinline]k\equiv\dfrac{1}{4\pi\varepsilon_0}=9\times 10^9\, N \cdot m^2 / C^2[/mathjaxinline].</p>
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<p><b> (Part b) </b> What if you made your capacitor out of thin sheets of foil held
[mathjaxinline]10 \, \mu m[/mathjaxinline] apart. Suppose
the foil sheets were [mathjaxinline] 2 \, cm[/mathjaxinline] wide and
[mathjaxinline] 1\,m[/mathjaxinline] long. Calculate the capacitance of
this system in [mathjaxinline]\mu F[/mathjaxinline].
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[mathjaxinline]C =[/mathjaxinline] </p>
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<p style="display:inline"> (in [mathjaxinline]\mu F[/mathjaxinline]) </p>
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<p><b> (Part c)</b> Suppose you connect a [mathjaxinline] 1.5 \, V[/mathjaxinline] battery to this second capacitor. What is the strength of the electric field inside the capacitor in Volts/meter?
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[mathjaxinline]|\vec{\mathbf{E}}| =[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">L12Q4: Capacitors Increase the Distance Between Plates</h2>
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Capacitors Increase the Distance Between Plates, Part 1
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<p><b class="bfseries">(Part a)</b> For each of these geometries of capacitor, how does the capacitance change if the distance between the plates increases (if [mathjaxinline]b[/mathjaxinline] goes to [mathjaxinline]2b[/mathjaxinline])? </p>
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a) Parallel plate capacitor (Figure 1): If [mathjaxinline]b[/mathjaxinline] changes to [mathjaxinline]2b[/mathjaxinline], the capacitance <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_checkpoint_w5_12_2_1">
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b) Cylindrical capacitor (Figure 2): If [mathjaxinline]b[/mathjaxinline] changes to [mathjaxinline]2b[/mathjaxinline], the capacitance <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 2" role="group"><div class="choicegroup capa_inputtype" id="inputtype_checkpoint_w5_12_3_1">
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c) Spherical capacitor (Figure 3): If [mathjaxinline]b[/mathjaxinline] changes to [mathjaxinline]2b[/mathjaxinline], the capacitance <div class="wrapper-problem-response" tabindex="-1" aria-label="Question 3" role="group"><div class="choicegroup capa_inputtype" id="inputtype_checkpoint_w5_12_4_1">
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Capacitors Increase the Distance Between Plates, Part 2
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<p><b class="bfseries">(Part b)</b> Assume that the parallel plate capacitor in Figure 1 has a capacitance [mathjaxinline]C_0[/mathjaxinline]. When the separation between plates [mathjaxinline]b[/mathjaxinline] is its capacitance is [mathjaxinline]C_{new}[/mathjaxinline]. Assume the plates to be large enough so we can neglect the edge effects. Calculate the ratio [mathjaxinline]C_{new}/C_0[/mathjaxinline] in terms of [mathjaxinline]b[/mathjaxinline], [mathjaxinline]A[/mathjaxinline] for the area of the plates, pi for [mathjaxinline]\pi[/mathjaxinline] and epsilon_0 for [mathjaxinline]\varepsilon _0[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]C_{new}/C_0 =[/mathjaxinline] </p>
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Capacitors Increase the Distance Between Plates, Part 3
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<p><b class="bfseries">(Part c)</b> The cylindrical capacitor in Figure 2 has a capacitance [mathjaxinline]C_0[/mathjaxinline]. When the radius of the outer shell, [mathjaxinline]b[/mathjaxinline], is doubled, keeping radius of the inner shell, [mathjaxinline]a[/mathjaxinline], fixed, the new capacitance is [mathjaxinline]C_{new}[/mathjaxinline]. Assume the shells to be very long with respect to [mathjaxinline]b[/mathjaxinline] so that we can neglect the edge effects. Calculate the ratio [mathjaxinline]C_{new}/C_0[/mathjaxinline] in terms of [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], [mathjaxinline]h[/mathjaxinline], pi for [mathjaxinline]\pi[/mathjaxinline], and epsilon_0 for [mathjaxinline]\varepsilon _0[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]C_{new}/C_0 =[/mathjaxinline] </p>
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Capacitors Increase the Distance Between Plates, Part 4
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<p><b class="bfseries">(Part d)</b> The spherical plate capacitor in Fgure 3 has a capacitance [mathjaxinline]C_0[/mathjaxinline]. If [mathjaxinline]b[/mathjaxinline], the radius of the outer shell is doubled, keeping the inner shell's radius, [mathjaxinline]a[/mathjaxinline], fixed, its capacitance is [mathjaxinline]C_{new}[/mathjaxinline]. Calculate the ratio [mathjaxinline]C_{new}/C_0[/mathjaxinline]. Express your answer in terms of [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], pi for [mathjaxinline]\pi[/mathjaxinline] and epsilon_0 for [mathjaxinline]\varepsilon _0[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]C_{new}/C_0 =[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_checkpoint_w5_12d_2_1" class="text-input-dynamath capa_inputtype inline textline">
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<h2 class="hd hd-2 unit-title">L12Q5: Energy Stored in a Capacitor</h2>
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Energy Stored in a Capacitor
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Let us consider how much energy it takes to charge a parallel plate capacitor starting from no charge up to a total charge of [mathjaxinline]+Q[/mathjaxinline] on the top plate and [mathjaxinline]-Q[/mathjaxinline] on the bottom plate. </p>
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<p><b class="bfseries">(Part a)</b> The way we will do this is to start with an uncharged capacitor and carry a little amount of charge [mathjaxinline]+dq[/mathjaxinline] from the bottom plate up to the top plate. We do this until we have a total charge of [mathjaxinline]+q[/mathjaxinline] on the top plate and [mathjaxinline]-q[/mathjaxinline] on the bottom plate. At this point in the process, what is the potential difference between these plates? Express your answer in terms of [mathjaxinline]q[/mathjaxinline] and [mathjaxinline]C[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\Delta V=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> The next little amount of charge [mathjaxinline]dq[/mathjaxinline] that we try to bring up from the negative to the positive plate with therefore be traveling across this potential difference and we will need to do a certain amount of work to bring that charge up across that potential difference. </p>
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What is the amount of work we need to do in order to bring this new little amount of charge [mathjaxinline]dq[/mathjaxinline] up to the positive plate? Write your answer in terms of [mathjaxinline]dq[/mathjaxinline], [mathjaxinline]q[/mathjaxinline], and [mathjaxinline]C[/mathjaxinline]? </p>
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<p style="display:inline">[mathjaxinline]dU=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> If we add up all this work, we can figure out the total amount of energy needed to charge up this capacitor (or equivalently how much energy is stored in the capacitor). If we start with no charge on the capacitor and end up with a total charge of [mathjaxinline]+Q[/mathjaxinline] on the top and [mathjaxinline]-Q[/mathjaxinline] on the bottom, how much energy is stored in the capacitor? Write your answer in terms of [mathjaxinline]Q[/mathjaxinline] and [mathjaxinline]C[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]U=[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">L12Q6: Capacitor Energy Scaling</h2>
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Capacitor Energy Scaling
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<p>Suppose a battery is connected to a parallel-plate capacitor with
capacitance [mathjaxinline]C[/mathjaxinline], and charges it up to a potential difference [mathjaxinline]V[/mathjaxinline]. Then the
battery is disconnected and the spacing between the capacitor plates is doubled. What is the change in each of the
following?</p>
<p><b> (Part a) </b> The volume</p>
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<option value="quadruples"> quadruples</option>
<option value="doubles"> doubles</option>
<option value="unchanged"> unchanged</option>
<option value="halves"> halves</option>
<option value="goes to one-fourth"> goes to one-fourth</option>
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<p><b> (Part b) </b> The electric field</p>
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<option value="quadruples"> quadruples</option>
<option value="doubles"> doubles</option>
<option value="unchanged"> unchanged</option>
<option value="halves"> halves</option>
<option value="goes to one-fourth"> goes to one-fourth</option>
</select>
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<p><b> (Part c) </b> The electric potential</p>
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<option value="quadruples"> quadruples</option>
<option value="doubles"> doubles</option>
<option value="unchanged"> unchanged</option>
<option value="halves"> halves</option>
<option value="goes to one-fourth"> goes to one-fourth</option>
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<p><b> (Part d) </b> The energy density</p>
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<option value="quadruples"> quadruples</option>
<option value="doubles"> doubles</option>
<option value="unchanged"> unchanged</option>
<option value="halves"> halves</option>
<option value="goes to one-fourth"> goes to one-fourth</option>
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<p><b> (Part e) </b> The amount of energy stored</p>
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<option value="doubles"> doubles</option>
<option value="unchanged"> unchanged</option>
<option value="halves"> halves</option>
<option value="goes to one-fourth"> goes to one-fourth</option>
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<p><b> (Part f) </b> The capacitance</p>
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<option value="quadruples"> quadruples</option>
<option value="doubles"> doubles</option>
<option value="unchanged"> unchanged</option>
<option value="halves"> halves</option>
<option value="goes to one-fourth"> goes to one-fourth</option>
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