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AP Physics 2 : Circuits

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Example Questions

Example Question #295 : Electricity And Magnetism

A dielectric is put between the plates of an isolated charged parallel plate capacitor. Which of the following statements is true?

Possible Answers:

The capacitance decreases

The potential difference decreases

The potential difference increases

The charge on the plates decreases

The charge on the plates increases

Correct answer:

The potential difference decreases

Explanation:

When a dielectric is added to a capacitor, the capacitance increases. In our problem, it says the system is charged and isolated, which means no charge will escape the system. Therefore the charge on the plates will stay the same.

The equation for capacitance is:

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

Since the charge stays the same, and the capacitance increases, the potential difference must decrease so that may increase.

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Example Question #6 : Properties Of Capacitors And Dielectrics

Suppose I have a uniform electric field within a parallel plate capacitor.

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

Given that the capacitance is $2.95*10^{-13}F$ , at what voltage difference is required for the capacitor to store $10^{-6}C$ of charge?

Possible Answers:

$3.05*10^{5} V$

3.4*10^6 V

$2.95*10^{-7} V$

6.8*10^6 V

Correct answer:

3.4*10^6 V

Explanation:

To determine this, we can use the formula:

$Q=C*\Delta V$

Where is charge stored, $\Delta V$ is voltage difference across a capacitor, and is capacitance.

Solving for $\Delta V$ ,

$\Delta V=\frac{Q}{C}=\frac{10^{-6}C}{2.95*10^{-13}F}=3.4*10^{6}V$

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Example Question #1 : Capacitors And Electric Fields

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

Possible Answers:

$10\frac{V}{m}$

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

$100\frac{V}{m}$

$1000\frac{V}{m}$

$1\frac{V}{m}$

Correct answer:

$1000\frac{V}{m}$

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:

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

Solving for the voltage:

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

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

V=Ed

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

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Example Question #2 : Capacitors And Electric Fields

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

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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

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

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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 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:

Increases

Stays constant

Decreases

We need to know

Correct answer:

Stays constant

Explanation:

Relevant equations:

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

Q=CV

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

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:

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

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:

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

Substitute C from the first equation:

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

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

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

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

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Example Question #1 : 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:

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

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

Stays constant

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

Correct answer:

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

Explanation:

Relevant equations:

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

Q=CV

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

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

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 #5 : Capacitors And Electric Fields

Lazy capacitor

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$V_1=V_{C1}$

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

$V_{C1}=E*d$

Combine these equations and solve for the electric field:

V_1=E*d

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

Convert mm to m and plug in values:

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

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

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Example Question #951 : Ap Physics 2

Lazy capacitor

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

Possible Answers:

31.9nC

25.4nC

50.6nC

48.8nC

27.7nC

Correct answer:

31.9nC

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 plugging in values:

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

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

Use the electric field in a capacitor equation:

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

Combine equations:

$E*\epsilon_0*A=q$

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

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

31.9nC=q

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Example Question #31 : Circuit Components

Lazy capacitor

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

$V_1=V_{C1}$

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:

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

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

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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 that are 2.5mm apart. If V_1 is equal to , determine the electric field between the plates.

Possible Answers:

$1500\frac{N}{C}$

$2000\frac{N}{C}$

$200\frac{N}{C}$

None of these

$2500\frac{N}{C}$

Correct answer:

$2000\frac{N}{C}$

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.

$V_1=V_{C1}$

Definition of voltage:

V=E*d

Combine equations:

$V_1=E_{C1}*d$

Solve for $E_{C1}$

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

Convert to meters and plug in values:

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

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

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AP Physics 2 : Circuits

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #295 : Electricity And Magnetism

Example Question #6 : Properties Of Capacitors And Dielectrics

Example Question #1 : Capacitors And Electric Fields

Example Question #2 : Capacitors And Electric Fields

Example Question #2 : Capacitors And Electric Fields

Example Question #1 : Capacitors And Electric Fields

Example Question #5 : Capacitors And Electric Fields

Example Question #951 : Ap Physics 2

Example Question #31 : Circuit Components

Example Question #1 : Capacitors And Electric Fields

All AP Physics 2 Resources

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