Capacitors and Electric Fields - AP Physics 2

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Question

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 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?

Answer

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:

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

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Question

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

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

Answer

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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Question

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?

Answer

Relevant equations:

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

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

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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Question

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?

Answer

Relevant equations:

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

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

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:

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

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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Question

In the given circuit, the capacitor is made of two parallel circular plates of radius that are apart. If is equal to , determine the electric field between the plates.

Answer

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:

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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Question

Consider the given diagram. If , each plate of the capacitor has surface area , and the plates at apart, determine the electric field between the plates.

Answer

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:

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

Convert mm to m and plug in values:

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

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

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Question

If , each plate of the capacitor has surface area , and the plates are apart, determine the electric field between the plates.

Answer

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.

$V_{C1}=Ed$

Combine these equations and solve for the electric field:

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

Convert mm to m and plug in values:

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

$E=4.810^4 \frac{N}{C}$

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Question

If , each plate of the capacitor has surface area , and the plates are apart, determine the excess charge on the positive plate.

Answer

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.

$V_{C1}=Ed$

Combine equations and solve for the electric field:

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

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:

$E\epsilon_0A=q$

Converting to and plug in values:

$600008.8510^{-12}*.06=q$

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Question

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

Answer

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

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

Combing equations and solving for

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

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

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Question

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

Answer

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

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

Combing equations and solving for

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

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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Question

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

Answer

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.

$V_{C1}=Ed$

Combining equations:

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:

$E\epsilon_0A=q$

Converting to and plugging in values:

$(6000)(8.8610^{-12})(410^{-4})=q$

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

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Question

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

Answer

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.

$V_{C1}=Ed$

Combining equations:

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:

$E\epsilon_0A=q$

Converting to and plugging in values:

$(4000)(8.8610^{-12})(410^{-4})=q$

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

Compare your answer with the correct one above

Question

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 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?

Answer

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:

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

Compare your answer with the correct one above

Question

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

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

Answer

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.

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

Compare your answer with the correct one above

Question

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?

Answer

Relevant equations:

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

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

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.

Compare your answer with the correct one above

Question

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?

Answer

Relevant equations:

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

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

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:

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

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

Compare your answer with the correct one above

Question

In the given circuit, the capacitor is made of two parallel circular plates of radius that are apart. If is equal to , determine the electric field between the plates.

Answer

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:

Combine equations:

$V_1=E_{C1}*d$

Solve for $E_{C1}$

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

Convert to meters and plug in values:

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

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

Compare your answer with the correct one above

Question

Consider the given diagram. If , each plate of the capacitor has surface area , and the plates at apart, determine the electric field between the plates.

Answer

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:

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

Convert mm to m and plug in values:

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

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

Compare your answer with the correct one above

Question

If , each plate of the capacitor has surface area , and the plates are apart, determine the electric field between the plates.

Answer

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.

$V_{C1}=Ed$

Combine these equations and solve for the electric field:

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

Convert mm to m and plug in values:

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

$E=4.810^4 \frac{N}{C}$

Compare your answer with the correct one above

Question

If , each plate of the capacitor has surface area , and the plates are apart, determine the excess charge on the positive plate.

Answer

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.

$V_{C1}=Ed$

Combine equations and solve for the electric field:

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

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:

$E\epsilon_0A=q$

Converting to and plug in values:

$600008.8510^{-12}*.06=q$

Compare your answer with the correct one above

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