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AP Physics C Electricity : Using Capacitor Equations

Study concepts, example questions & explanations for AP Physics C Electricity

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Example Questions

Example Question #1 : Capacitors

You are hired to make a capacitor out of two parallel metal sheets. If someone wanted you to make a thin capacitor of $700\mu F$ out of those metal sheets, and the sheets needed to be 0.5mm apart, what area do the two metal plates need to be?

$\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Possible Answers:

6.21*10^6m^2

3.95*10^4m^2

5.47*10^8m^2

$3.95*10^{-4}m^2$

2.77*10^3m^2

Correct answer:

3.95*10^4m^2

Explanation:

For parallel plate capacitors, the equation is $C=\frac{\epsilon_0A}{d}$ .

Solve for .

$A=\frac{dC}{\epsilon_0}$

Now we can plug in our given values.

$A=\frac{(0.5*10^{-3})(700*10^{-6})}{8.85*10^{-12}\frac{\text{C}^2}{\text{Nm}^2}}$

$A=3.95*10^4\ \text{m}^2$

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Example Question #1 : Using Capacitor Equations

The plates of a parallel plate capacitor are 3mm apart and 5m^2 in area. A potential difference of 1000V is applied across the capacitor. Find the capacitance.

$\epsilon_0=8.85*10^{-12} \frac{F}{m}$

Possible Answers:

$1.5*10^{-8}F$

$1.5*10^{-9}F$

$1.5*10^{-7}F$

$1.5*10^{-10}F$

Correct answer:

$1.5*10^{-8}F$

Explanation:

Capacitance is related to plate area and distance by the equation $C=\epsilon_0\frac{A}{d}$ .

Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.

$C=8.85* 10^{-12} \frac{F}{m}(\frac{5 m^2}{3*10^{-3}m})=8.85* 10^{-12} \frac{F}{m}(1.67*10^{3}m)=1.5*10^{-8}F$

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Example Question #1 : Electric Circuits

The plates of a parallel plate capacitor are 3mm apart and 5m^2 in area. A potential difference of 1000V is applied across the capacitor. Compute the charge on each plate.

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

Possible Answers:

$1.5*10^{-7}C$

$1.5*10^{-6}C$

$1.5*10^{-8}C$

$1.5*10^{-5}C$

Correct answer:

$1.5*10^{-5}C$

Explanation:

Charge on a capacitor is given by the equation Q=CV . We know that the voltage is 1000V , but we need to determine the capacitance based off of the area and distance between the plates.

$C=\epsilon_0\frac{A}{d}$

$C=(8.85*10^{-12}\frac{F}{m})(\frac{5m^2}{3*10^{-3}m})$

$C=(8.85*10^{-12}\frac{F}{m})(1.67*10^{3}m)=1.5*10^{-8}F$

We can plug this value into the equation for charge.

$Q=CV=(1.5*10^{-8}F)(1000V)$

$Q=1.5*10^{-5}C$

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Example Question #2 : Electric Circuits

The plates of a parallel plate capacitor are 3mm apart. A potential difference of 1000V is applied across the capacitor. Compute the magnitude of the generated electric field.

$\epsilon_0=8.85*10^{-12}\frac{F}{m}$

Possible Answers:

$3.33*10^3\frac{N}{C}$

$3.33*10^4\frac{N}{C}$

$3.33*10^5\frac{N}{C}$

$3.33*10^6\frac{N}{C}$

Correct answer:

$3.33*10^5\frac{N}{C}$

Explanation:

The electric field given by a capacitor is given by the formula $E=\frac{Q}{\epsilon_0 A}$ . We do not have these variables, so we will have to adjust the equation.

$Q=CV\rightarrow E=\frac{CV}{\epsilon_0A}$

The capacitance can be determined by the area of the plates and the distance between them.

$C=\epsilon_0\frac{A}{d}\rightarrow E=\frac{\epsilon_0\frac{A}{d}V}{\epsilon_0A}$

This equation simplifies to $E=\frac{V}{d}$ , allowing us to solve using the values given in the question.

$E=\frac{1000V}{3*10^{-3}m}=3.33*10^{5}\frac{N}{C}$

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Example Question #1 : Electric Circuits

If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

Possible Answers:

$C_{eq}=C$

$C_{eq}=\frac{C}{2}$

$C < C_{eq}< 2C$

$C_{eq}= 2C$

$\frac{C}{2} < C_{eq} < C$

Correct answer:

$C_{eq}=\frac{C}{2}$

Explanation:

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

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Example Question #1 : Using Capacitor Equations

Two parallel conducting plates each have an area of and are separated by a distance, . One plate has a charge evenly distributed across it, and the other has a charge of . A proton (charge ) is initially held near the positive plate, then released such that it accelerates towards the negative plate. How much kinetic energy has the proton gained in this process?

Possible Answers:

$\frac{eQd}{\epsilon _oA}$

$\frac{eQ\epsilon _o A}{d}$

$\frac{e \epsilon _o A}{Qd}$

$\frac{eQ}{\epsilon _o A d}$

$\frac{eQ \epsilon _o A}{d^2}$

Correct answer:

$\frac{eQd}{\epsilon _oA}$

Explanation:

Relevant equations:

$W = q\Delta V$

$W = \Delta KE$

$C = \frac{Q}{V}$

$C = \frac{\epsilon _o A}{d}$

According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.

1. Find an expression for potential difference, , in terms of and , by setting the two capacitance equations equal to each other:

$\frac{Q}{V}=\frac{\epsilon _o A}{d}$

$V = \frac{Qd}{\epsilon _o A}$

2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):

$W = \frac {eQd}{\epsilon _o A} = \Delta K$

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Example Question #211 : Ap Physics C

A $5 \mu F$ and $10 \mu F$ capacitor are connected in series with a 12V battery. What is the magnitude of the charge on one plate of the $5 \mu F$ capacitor?

Possible Answers:

$180 \mu C$

$60 \mu C$

$10 \mu C$

$2.4 \mu C$

$40 \mu C$

Correct answer:

$40 \mu C$

Explanation:

Relevant equations:

Q = CV

$\frac{1}{C_{eq, series}} = \frac{1}{C_1} + \frac{1}{C_2}$

First, find the equivalent capacitance of the whole circuit:

$\frac{1}{C_{eq,series}}= \frac{1}{5} + \frac{1}{10} = \frac{3}{10}$

$C_{eq, series} = \frac{10}{3} \mu F$

Use this equivalent capacitance to find the total charge:

$Q = CV = (\frac{10}{3} \mu F) * (12 V) = 40 \mu C$

Capacitors in series always have the same charge on each unit, so the charge on the $5 \mu F$ capacitor is also $40 \mu C$ .

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AP Physics C Electricity : Using Capacitor Equations

Study concepts, example questions & explanations for AP Physics C Electricity

All AP Physics C Electricity Resources

Example Questions

Example Question #1 : Capacitors

Example Question #1 : Using Capacitor Equations

Example Question #1 : Electric Circuits

Example Question #2 : Electric Circuits

Example Question #1 : Electric Circuits

Example Question #1 : Using Capacitor Equations

Example Question #211 : Ap Physics C

All AP Physics C Electricity Resources

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