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AP Physics 2 : Capacitors and Electric Fields

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Example Questions

Example Question #1 : Capacitors And Electric Fields

A $3.0\: \mu F$ $\displaystyle 3.0\: \mu F$ parallel-plate capacitor is connected to a constant voltage source. If the distance between the plates of this capacitor is $4\:mm$ $\displaystyle 4\:mm$ and the capacitor holds a charge of $12.0\: \mu C$ $\displaystyle 12.0\: \mu C$ , what is the value of the electric field between the plates of this capacitor?

Possible Answers:

$10\frac{V}{m}$ $\displaystyle 10\frac{V}{m}$

An electric field does not exist between the plates of a parallel-plate capacitor

$1000\frac{V}{m}$ $\displaystyle 1000\frac{V}{m}$

$1\frac{V}{m}$ $\displaystyle 1\frac{V}{m}$

$100\frac{V}{m}$ $\displaystyle 100\frac{V}{m}$

Correct answer:

$1000\frac{V}{m}$ $\displaystyle 1000\frac{V}{m}$

Explanation:

To solve this problem, we first need to solve for the voltage across the capacitor. For this, we'll need the formula for capacitance:

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

Solving for the voltage:

$V=\frac{Q}{C}=\frac{12.0\cdot 10^{-6}C}{3.0\cdot 10^{-6}F}=4.0V$ $\displaystyle V=\frac{Q}{C}=\frac{12.0\cdot 10^{-6}C}{3.0\cdot 10^{-6}F}=4.0V$

Now that we have the voltage, we can make use of the following equation to solve for the electric field:

V=Ed $\displaystyle V=Ed$

$E=\frac{V}{d}=\frac{4.0V}{4\cdot 10^{-3}m}=1000\frac{V}{m}$ $\displaystyle E=\frac{V}{d}=\frac{4.0V}{4\cdot 10^{-3}m}=1000\frac{V}{m}$

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Example Question #2 : Capacitors And Electric Fields

A set of parallel plate capacitors with a surface area of 12m^2 $\displaystyle 12m^2$ has a total amount of charge equal to $\displaystyle 4C$ . What is the electric field between the plates?

$\epsilon=8.85\cdot 10^{-12}\frac{F}{m}$ $\displaystyle \epsilon=8.85\cdot 10^{-12}\frac{F}{m}$

Possible Answers:

$8.12\cdot 10^{10}\frac{N}{C}$ $\displaystyle 8.12\cdot 10^{10}\frac{N}{C}$

$3.77\cdot 10^{10}\frac{N}{C}$ $\displaystyle 3.77\cdot 10^{10}\frac{N}{C}$

$6.67\cdot 10^{9}\frac{N}{C}$ $\displaystyle 6.67\cdot 10^{9}\frac{N}{C}$

$5.34\cdot 10^{10}\frac{N}{C}$ $\displaystyle 5.34\cdot 10^{10}\frac{N}{C}$

$8.54\cdot 10^{9}\frac{N}{C}$ $\displaystyle 8.54\cdot 10^{9}\frac{N}{C}$

Correct answer:

$3.77\cdot 10^{10}\frac{N}{C}$ $\displaystyle 3.77\cdot 10^{10}\frac{N}{C}$

Explanation:

The equation for the electric field between two parallel plate capacitors is:

$E=\frac{\sigma}{\epsilon}$ $\displaystyle E=\frac{\sigma}{\epsilon}$

Sigma is the charge density of the plates, which is equal to:

$\sigma = \frac{Q}{A}$ $\displaystyle \sigma = \frac{Q}{A}$

We are given the area and total charge, so we use them to find the charge density.

$\sigma&=\frac{4}{12}$ $\displaystyle \sigma&=\frac{4}{12}$

$\sigma= 0.3334\frac{C}{m^2}$ $\displaystyle \sigma= 0.3334\frac{C}{m^2}$

Now that we have the charge density, divide it by the vacuum permittivity to find the electric field.

$E=\frac{0.3334}{8.85\times10^{-12}}=3.77\times10^{10}\frac{N}{C}$ $\displaystyle E=\frac{0.3334}{8.85\times10^{-12}}=3.77\times10^{10}\frac{N}{C}$

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Example Question #3 : Capacitors And Electric Fields

Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is not connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the electric field, E, stored in the capacitor increase, decrease or stay the same?

Possible Answers:

Decreases

Stays constant

We need to know

Increases

Correct answer:

Stays constant

Explanation:

Relevant equations:

$C=\frac{\varepsilon_{o}A}{D}$ $\displaystyle C=\frac{\varepsilon_{o}A}{D}$

Q=CV $\displaystyle Q=CV$

$U=\frac{1}{2}\frac{Q^{2}}{C}$ $\displaystyle U=\frac{1}{2}\frac{Q^{2}}{C}$

Considering we are dealing with regions of charge instead of a point charge (or a charge that is at least spherical symmetric), it would be wise to consider using the definition of the electric field here:

$E=\frac{V}{D}$ $\displaystyle E=\frac{V}{D}$

We are only considering magnitude so the direction of the electric field is not of concern. Considering a battery is not connected, we can't claim the V is constant. So we use the second equation to substitute for V in the above equation:

$E=\frac{Q}{CD}$ $\displaystyle E=\frac{Q}{CD}$

Substitute C from the first equation:

$E=\frac{Q\varepsilon _{o}D}{AD}$ $\displaystyle E=\frac{Q\varepsilon _{o}D}{AD}$

We see here that D actually cancels. This would not have been obvious without the previous substitution. Final result:

$E=\frac{\varepsilon _{o}Q}{A}$ $\displaystyle E=\frac{\varepsilon _{o}Q}{A}$

These 3 quantities are static in this situation so E does not change.

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Example Question #2 : Capacitors And Electric Fields

Imagine a capacitor with a magnitude of charge Q on either plate. This capacitor has area A, separation distance D, and is connected to a battery of voltage V. If some external agent pulls the capacitor apart such that D doubles, did the electric field, E, stored in the capacitor increase, decrease or stay the same?

Possible Answers:

Stays constant

Decreases by exactly $\frac{1}{4}$ $\displaystyle \frac{1}{4}$

Increases by exactly $\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Decreases by exactly $\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Correct answer:

Decreases by exactly $\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Explanation:

Relevant equations:

$C=\frac{\varepsilon_{o}A}{D}$ $\displaystyle C=\frac{\varepsilon_{o}A}{D}$

Q=CV $\displaystyle Q=CV$

$U=\frac{1}{2}\frac{Q^{2}}{C}$ $\displaystyle U=\frac{1}{2}\frac{Q^{2}}{C}$

$E=\frac{V}{D}$ $\displaystyle E=\frac{V}{D}$

The battery maintains a constant V so we can directly relate E and D. Since D is doubled, E will be halved.

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Example Question #3 : Capacitors And Electric Fields

Lazy capacitor

If V_1=24V $\displaystyle V_1=24V$ , each plate of the capacitor has surface area $\displaystyle 240 cm^2$ , and the plates are .5mm $\displaystyle .5mm$ apart, determine the electric field between the plates.

Possible Answers:

$1.6*10^4 \frac{N}{C}$ $\displaystyle 1.6*10^4 \frac{N}{C}$

$4.8*10^4 \frac{N}{C}$ $\displaystyle 4.8*10^4 \frac{N}{C}$

$4.0*10^4 \frac{N}{C}$ $\displaystyle 4.0*10^4 \frac{N}{C}$

$5.5*10^4 \frac{N}{C}$ $\displaystyle 5.5*10^4 \frac{N}{C}$

$6.6*10^4 \frac{N}{C}$ $\displaystyle 6.6*10^4 \frac{N}{C}$

Correct answer:

$4.8*10^4 \frac{N}{C}$ $\displaystyle 4.8*10^4 \frac{N}{C}$

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$ $\displaystyle V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$ $\displaystyle V_{C1}=E*d$

Combine these equations and solve for the electric field:

V_1=E*d $\displaystyle V_1=E*d$

$\frac{V_1}{d}=E$ $\displaystyle \frac{V_1}{d}=E$

Convert mm to m and plug in values:

$\frac{24}{.0005}=E$ $\displaystyle \frac{24}{.0005}=E$

$E=4.8*10^4 \frac{N}{C}$ $\displaystyle E=4.8*10^4 \frac{N}{C}$

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Example Question #4 : Capacitors And Electric Fields

Lazy capacitor

If V_1=60V $\displaystyle V_1=60V$ , each plate of the capacitor has surface area $\displaystyle 600 cm^2$ , and the plates are 1mm $\displaystyle 1mm$ apart, determine the excess charge on the positive plate.

Possible Answers:

50.6nC $\displaystyle 50.6nC$

31.9nC $\displaystyle 31.9nC$

25.4nC $\displaystyle 25.4nC$

48.8nC $\displaystyle 48.8nC$

27.7nC $\displaystyle 27.7nC$

Correct answer:

31.9nC $\displaystyle 31.9nC$

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$ $\displaystyle V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$ $\displaystyle V_{C1}=E*d$

Combine equations and solve for the electric field:

V_1=E*d $\displaystyle V_1=E*d$

$\frac{V_1}{d}=E$ $\displaystyle \frac{V_1}{d}=E$

Convert mm to m and plugging in values:

$\frac{60V}{.001}=E$ $\displaystyle \frac{60V}{.001}=E$

$60000\frac{N}{C}=E$ $\displaystyle 60000\frac{N}{C}=E$

Use the electric field in a capacitor equation:

$E=\frac{q}{A*\epsilon_0}$ $\displaystyle E=\frac{q}{A*\epsilon_0}$

Combine equations:

$E*\epsilon_0*A=q$ $\displaystyle E*\epsilon_0*A=q$

Converting cm^2 $\displaystyle cm^2$ to m^2 $\displaystyle m^2$ and plug in values:

$60000*8.85*10^{-12}*.06=q$ $\displaystyle 60000*8.85*10^{-12}*.06=q$

$\displaystyle 31.9nC=q$

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Example Question #5 : Capacitors And Electric Fields

Lazy capacitor

Consider the given diagram. If V_1=4V $\displaystyle V_1=4V$ , each plate of the capacitor has surface area $\displaystyle 100 cm^2$ , and the plates at 1mm $\displaystyle 1mm$ apart, determine the electric field between the plates.

Possible Answers:

$2000\frac{N}{C}=E$ $\displaystyle 2000\frac{N}{C}=E$

$7000\frac{N}{C}=E$ $\displaystyle 7000\frac{N}{C}=E$

$9000\frac{N}{C}=E$ $\displaystyle 9000\frac{N}{C}=E$

$3333\frac{N}{C}=E$ $\displaystyle 3333\frac{N}{C}=E$

$4000\frac{N}{C}=E$ $\displaystyle 4000\frac{N}{C}=E$

Correct answer:

$4000\frac{N}{C}=E$ $\displaystyle 4000\frac{N}{C}=E$

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor.

$V_1=V_{C1}$ $\displaystyle V_1=V_{C1}$

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

$V_{C1}=E*d$ $\displaystyle V_{C1}=E*d$

Combine equations and solve for the electric field:

V_1=E*d $\displaystyle V_1=E*d$

$\frac{V_1}{d}=E$ $\displaystyle \frac{V_1}{d}=E$

Convert mm to m and plug in values:

$\frac{V_1}{d}=E$ $\displaystyle \frac{V_1}{d}=E$

$4000\frac{N}{C}=E$ $\displaystyle 4000\frac{N}{C}=E$

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Example Question #1 : Capacitors And Electric Fields

Lazy capacitor

In the given circuit, the capacitor is made of two parallel circular plates of radius .3cm $\displaystyle .3cm$ that are 2.5mm $\displaystyle 2.5mm$ apart. If V_1 $\displaystyle V_1$ is equal to $\displaystyle 5V$ , determine the electric field between the plates.

Possible Answers:

None of these

$2000\frac{N}{C}$ $\displaystyle 2000\frac{N}{C}$

$200\frac{N}{C}$ $\displaystyle 200\frac{N}{C}$

$1500\frac{N}{C}$ $\displaystyle 1500\frac{N}{C}$

$2500\frac{N}{C}$ $\displaystyle 2500\frac{N}{C}$

Correct answer:

$2000\frac{N}{C}$ $\displaystyle 2000\frac{N}{C}$

Explanation:

Since it is the only element in the circuit besides the source, the voltage drop across the capacitor must be equal to the voltage gain in the capacitor.

$V_1=V_{C1}$ $\displaystyle V_1=V_{C1}$

Definition of voltage:

V=E*d $\displaystyle V=E*d$

Combine equations:

$V_1=E_{C1}*d$ $\displaystyle V_1=E_{C1}*d$

Solve for $E_{C1}$ $\displaystyle E_{C1}$

$\frac{V}{d}=E_{C1}$ $\displaystyle \frac{V}{d}=E_{C1}$

Convert to meters and plug in values:

$\frac{5}{.0025}=E_{C1}$ $\displaystyle \frac{5}{.0025}=E_{C1}$

$2000\frac{N}{C}=E_{C1}$ $\displaystyle 2000\frac{N}{C}=E_{C1}$

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Example Question #7 : Capacitors And Electric Fields

A parallel plate capacitor with a separation of $.5\textup{ mm}$ $\displaystyle .5\textup{ mm}$ and surface area $150\textup{ cm}^2$ $\displaystyle 150\textup{ cm}^2$ is in series with a $9\textup{ V}$ $\displaystyle 9\textup{ V}$ battery. How will the electric field in the capacitor change if the separation is doubled?

Possible Answers:

Halved

It will stay the same

Tripled

Quadrupled

Doubled

Correct answer:

Halved

Explanation:

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

V_b=V_c $\displaystyle V_b=V_c$

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

V_c=E*d $\displaystyle V_c=E*d$

Combing equations and solving for $\displaystyle E$

$E=\frac{V_b}{d}$ $\displaystyle E=\frac{V_b}{d}$

From this, it can be seen that doubling the separation will halve the electric field.

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Example Question #1 : Capacitors And Electric Fields

Possible Answers:

Impossible to determine

No change

Tripled

Doubled

Halved

Correct answer:

Doubled

Explanation:

The voltage drop through the capacitor needs to be equal to the voltage of the battery.

V_b=V_c $\displaystyle V_b=V_c$

The voltage drop of a parallel plate capacitor is equal to the internal electric field times the distance between them.

V_c=E*d $\displaystyle V_c=E*d$

Combing equations and solving for $\displaystyle E$

$E=\frac{V_b}{d}$ $\displaystyle E=\frac{V_b}{d}$

From this, it can be seen that doubling the voltage of the battery will doubled the electric field inside the capacitor.

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AP Physics 2 : Capacitors and Electric Fields

Study concepts, example questions & explanations for AP Physics 2

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