What Property of Objects Is Best Measured by Their Capacitance?

Introduction to Capacitance

Part A

What property of objects is best measured by their capacitance?

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ability to store charge
Correct

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Capacitance is a measure of the ability of a system of two conductors to store electric charge and energy. Capacitance is defined as . This ratio remains constant as long as the system retains its geometry and the amount of dielectric does not change. Capacitors are special devices designed to combine a large capacitance with a small size. However, any pair of conductors separated by a dielectric (or vacuum) has some capacitance. Even an isolated electrode has a small capacitance. That is, if a charge is placed on it, its potential with respect to ground will change, and the ratio is its capacitance .

Part B

Assume that charge is placed on the top plate, and is placed on the bottom plate. What is the magnitude of the electric field between the plates?

Express in terms of and other quantities given in the introduction, in addition to and any other constants needed.

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=(q/(epsilon_0*A))Correct

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Part C

What is the voltage between the plates of the capacitor?

Express in terms of the quantities given in the introduction and any required physical constants.

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=((q*d)/(epsilon_0*A))Correct

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Part D

Now find the capacitance of the parallel-plate capacitor.

Express in terms of quantities given in the introduction and constants like .

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=epsilon_0*(A/d)Correct

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You have derived the general formula for the capacitance of a parallel-plate capacitor with plate area and plate separation . It is worth remembering.

Part E

Consider an air-filled charged capacitor. How can its capacitance be increased?

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Decrease the spacing between the plates of the capacitor.
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Part F

Consider a charged parallel-plate capacitor. Which combination of changes would quadruple its capacitance?

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Halve the plate separation; double the plate area.
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Finding the Capacitance

Part A

What is the capacitance of the second capacitor?

Express your answer in terms of , , and constants.

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=(K*C_1)/(K-1)Correct

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As a check that this result makes sense, note that if , which corresponds to the first capacitor being replaced by a conductor, then . In the limit as approaches 1 (but is still greater than 1), becomes very large, effectively making the second capacitor a conductor. This is also expected, since if , then no change was made to the first capacitor.

Force between Capacitor Plates

Part A

Find , the magnitude of the force each plate experiences due to the other plate as a function of , the potential drop across the capacitor.

Express your answer in terms of given quantities and .

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=(A*epsilon_0*V^2)/(2*d^2)Correct

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Capacitors with Partial Dielectrics

Part A

The horizontal capacitor is filled halfway with a material that has dielectric constant . What fraction of the area of the vertical capacitor should be filled (as shown in the figure) with the same dielectric so that the two capacitors have equal capacitance?

Express your answer in terms of .

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=1/(1+K)Correct

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Energy of a Capacitor

Part A

Find , the the energy dissipated in the resistor.

Express your answer in terms of and other given quantities.

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=U/KCorrect

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Can you explain where the rest of the energy went?

Part B

Consider the same situation as in the previous part, except that the charging battery remains connected while the dielectric is inserted. The battery is then disconnected and the capacitor discharged. For this situation, what is , the energy dissipated in the resistor?

Express your answer in terms of and other given quantities.

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=U*KCorrect

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Can you explain where the extra energy came from? (Answer below)

In this case, the extra energy comes from the battery which provides some extra charge, to maintain the potential difference across the plates.

The first case is a little more involved. The dielectric experiences an attractive force as it is inserted into the capacitor. Therefore it would either gain kinetic energy, or work would have to be done against any resistive forces on it that prevent it from gaining speed. This is where the the energy goes. You could actually calculate the force on the dielectric as , where is the energy in the capacitor as a function of the length of the dielectric inserted into the capacitor.