<p>A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm^2, pl
ID: 2006170 • Letter: #
Question
<p>A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm^2, plate separation d = 10.0 mm and dielectric constant K = 5.00. The capacitor is connected to a battery that creates a constant voltage V = 5.00 V. Throughout the problem, E(not) = 8.85×10−12 C^2/N c* m^2.</p><p> </p>
<div class="partText">Find the energy <img title="U_1" src="http://session.masteringphysics.com/render?var=U_1" alt="U_1" align="middle" /> of the dielectric-filled capacitor.</div>
<div class="partInstructions">Express your answer numerically in joules.</div>
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<div class="answer"><span class="solveFor">  <img title="U_1" src="http://session.masteringphysics.com/render?var=U_1" alt="U_1" align="middle" />  =</span><span class="solution"><span> </span></span><span class="units"> _____ <img src="http://session.masteringphysics.com/render?tex=%5Crm+J" alt=" m J" align="middle" /> </span></div>
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<div id="part_id5760002" class="partVisible">Part B</div>
<div class="partText">The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy <img title="U_2" src="http://session.masteringphysics.com/render?var=U_2" alt="U_2" align="middle" /> of the capacitor at the moment when the capacitor is half-filled with the dielectric.</div>
<div class="partInstructions">Express your answer numerically in joules.</div>
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<div><span class="units">  _______ <img src="http://session.masteringphysics.com/render?tex=%5Crm+J" alt=" m J" align="middle" /> </span></div>
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<div id="part_id5761460" class="partVisible">Part C</div>
<div class="partText">The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, <img title="U_3" src="http://session.masteringphysics.com/render?var=U_3" alt="U_3" align="middle" />.</div>
<div class="partInstructions">Express your answer numerically in joules.</div>
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<div><span class="solveFor">  <img title="U_3" src="http://session.masteringphysics.com/render?var=U_3" alt="U_3" align="middle" />  =</span></div>
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<div><span class="units">  ________ <img src="http://session.masteringphysics.com/render?tex=%5Crm+J" alt=" m J" align="middle" /> </span></div>
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<div id="part_id5762394" class="partVisible">Part D</div>
<div class="partText">In the process of removing the remaining portion of the dielectric from the disconnected capacitor, how much work <img title="W" src="http://session.masteringphysics.com/render?var=W" alt="W" align="middle" /> is done by the external agent acting on the dielectric?</div>
<div class="partInstructions">Express your answer numerically in joules.</div>
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<div><span class="units">  __________ <img src="http://session.masteringphysics.com/render?tex=%5Crm+J" alt=" m J" align="middle" /> </span></div>
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Explanation / Answer
Given that Area of the capacitor is A = 30.0 cm^2 seperation distance is d = 10.0 mm dielectric constant is k = 5.00 Voltage is V = 5.00 V a ) Energy stored in a capacitor when the dielectric filled E_1 = 1/2 CV^2 capacitance C = k _0 A / d = 5.00 *8.85*10^-12 * 30*10^-4 m^2 / 10*10^-3 m = 13.2*10^-12 F E_1 = 1/2 CV^2 = 1/2 * 13.2*10^-12 F * 25 V = 165*10^-12 J b ) when the dielectric is half filled C = k /2 _0 A / d = 5.00/2 *8.85*10^-12 * 30*10^-4 m^2 / 10*10^-3 m = 6.63*10^-12 F E_2 = 1/2 CV^2 = 1/2 * 6.63*10^-12 F * 25 V = 82.8*10^-12 J = 1/2 * 6.63*10^-12 F * 25 V = 82.8*10^-12 J c ) When the dielecric is moved away C = _0 A / d = 8.85*10^-12 * 30*10^-4 m^2 / 10*10^-3 m = 2.6*10^-12 F = 8.85*10^-12 * 30*10^-4 m^2 / 10*10^-3 m = 2.6*10^-12 F E_3 = 1/2 CV^2 = 1/2 * 2.6*10^-12 F * 25 V = 33.1*10^-12 J d ) Work done when the dielectric is removed W = 1/2 CV^2 = 33.1*10^-12 J = 1/2 * 2.6*10^-12 F * 25 V = 33.1*10^-12 J d ) Work done when the dielectric is removed W = 1/2 CV^2 = 33.1*10^-12 JRelated Questions
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