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AP Physics 2 : Capacitors and Electric Fields

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Example Questions

Example Question #11 : Capacitors And Electric Fields

A capacitor is placed in series with a battery. The plates are 1mm apart and each have a 4cm^2 surface area. Determine the charge on the positive plate.

Possible Answers:

None of these

$4.54*10^{-11}C$

$2.13*10^{-11}C$

$8.44*10^{-11}C$

$3.95*10^{-11}C$

Correct answer:

$2.13*10^{-11}C$

Explanation:

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

$V_1=V_{C1}$

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

$V_{C1}=E*d$

Combining equations:

V_1=E*d

Solving for :

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

Converting to and plugging in values:

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

$6000\frac{N}{C}=E$

Using the electric field in a capacitor equation:

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

Combining equations:

$E*\epsilon_0*A=q$

Converting cm^2 to m^2 and plugging in values:

$(6000)*(8.86*10^{-12})*(4*10^{-4})=q$

$q=2.13*10^{-11}C$

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Example Question #311 : Electricity And Magnetism

A capacitor is placed in series with three parallel batteries. The plates are 1.5mm apart and each have a 10cm^2 surface area. Determine the charge on the positive plate.

Possible Answers:

$8.44*10^{-11}C$

None of these

$3.08*10^{-11}C$

$5.55*10^{-11}C$

$1.42*10^{-11}C$

Correct answer:

$1.42*10^{-11}C$

Explanation:

The voltage rise through the source must be the same as the drop through the capacitor. Since the batteries all have the same potential and are in parallel, the potential will be the same as if there was just one.

$V_1=V_{C1}$

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

$V_{C1}=E*d$

Combining equations:

V_1=E*d

Solving for E:

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

Converting to and plugging in values:

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

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

Using the electric field in a capacitor equation:

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

Combining equations:

$E*\epsilon_0*A=q$

Converting cm^2 to m^2 and plugging in values:

$(4000)*(8.86*10^{-12})*(4*10^{-4})=q$

$q=1.42*10^{-11}C$

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All AP Physics 2 Resources

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AP Physics 2 : Capacitors and Electric Fields

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #11 : Capacitors And Electric Fields

Example Question #311 : Electricity And Magnetism

All AP Physics 2 Resources

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