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AP Physics 2 : Capacitors and Capacitance

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Example Questions

Example Question #11 : Capacitors And Capacitance

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, how did the capacitance change? Also, do we need to use the information provided about the battery?

Possible Answers:

Increase, yes

Decrease, yes

Increase, no

Decrease, no

Correct answer:

Decrease, no

Explanation:

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

As D increases, C will decrease

Does the battery matter? No. Capacitance is a "geometric" quantity. This means that it can be determined solely by physical parameters like the area and separation distance. The charge on the plates, voltage difference, electric field and any other quantity you could think of does not influence the capacitance other than A or D.

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Example Question #112 : Circuits

Lazy capacitor

Consider the given diagram. If V_1=4 V , each plate of the capacitor has surface area 100 cm^2 , and the plates are 0.1mm apart, determine the number of excess electrons on the negative plate.

$e=1.6*10^{-19}C$

Possible Answers:

$n=2.21*10^{9}$

None of these

$n=5.85*10^{9}$

$n=5.15*10^{9}$

$n=3.33*10^{9}$

Correct answer:

$n=2.21*10^{9}$

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.

$V_{C1}=E*d$

Combine equations and find the electric field:

V_1=E*d

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

Convert mm to m and plugg in values:

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

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

Use the electric field in a capacitor equation:

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

$E*\epsilon_0*A=q$

Convert cm^2 to m^2 and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

$3.54*10^{-10}=q$

The magnitude of total charge on the positive plate is equal to the total charge on the negative plate, so to find the number of excess elections:

$\frac{3.54*10^{-10}}{1.6*10^{-19}}=n$

$n=2.21*10^{9}$

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Example Question #11 : Circuit Components

Lazy capacitor

Consider the given diagram. If V_1=4 V , each plate of the capacitor has surface area 100 cm^2 , and the plates at 1mm apart, determine the excess charge on the positive plate.

Possible Answers:

$4.44*10^{-10}C$

$1.11*10^{-10}C$

$9.26*10^{-10}C$

$3.54*10^{-10}C$

$6.11*10^{-10}C$

Correct answer:

$3.54*10^{-10}C$

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.

$V_{C1}=E*d$

Combine equations and solve for the electric field:

V_1=E*d

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

Convert mm to m and plug in values:

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

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

Using the electric field in a capacitor equation:

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

Rearrange to solve for the charge:

$E*\epsilon_0*A=q$

Convert cm^2 to m^2 and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

$3.54*10^{-10}C=q$

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Example Question #14 : Circuit Components

Suppose I have a uniform electric field within a parallel plate capacitor with field strength of $E=41\frac{N}{C}$ .

Suppose the capacitor's plates are .5m in length and .001m in width, and the space between the plates is 1.5cm .

What is the voltage difference across the capacitor?

Possible Answers:

.615V

61.5V

.475V

5.5V

Correct answer:

.615V

Explanation:

Voltage difference $\Delta V$ is given by

$\Delta V=E*d$ , where is the electric field strength and is the distance between the two plates.

For this problem,

$\Delta V=(41*.015)V= .615V$

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Example Question #15 : Circuit Components

Suppose I have a uniform electric field within a parallel plate capacitor with field strength of $E=41\frac{N}{C}$ .

Suppose the capacitor's plates are .5m in length and .001m in width, and the space between the plates is 1.5cm .

Determine the capacitance of this system given that the space in between is a vacuum, and that the permittivity of empty space $\epsilon_0$ is $8.85*10^{-12}\frac{F}{m}$ .

Possible Answers:

$C=2.95*10^{-12}F$

$C=2.95*10^{-13}F$

$C=3.38*10^{-6}F$

$C=3.37*10^{6}F$

Correct answer:

$C=2.95*10^{-13}F$

Explanation:

The formula for capacitance is given by:

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

Where $\kappa$ is dielectric strength, is distance between plates, $\epsilon _0$ is permittivity of empty space, and is cross sectional area.

To determine , we do

$A=(.5*.001)m^2=5*10^{-4}m^2$

$\kappa=1$ because between the plates is a vacuum.

d=.015m

Putting it all together,

$C=2.95*10^{-13}F$

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AP Physics 2 : Capacitors and Capacitance

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #11 : Capacitors And Capacitance

Example Question #112 : Circuits

Example Question #11 : Circuit Components

Example Question #14 : Circuit Components

Example Question #15 : Circuit Components

All AP Physics 2 Resources

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