Capacitors and Capacitance - AP Physics 2

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Question

If the maximum amount of charge held by a capacitor at a voltage of 12V is 36C, what is the capacitance of this capacitor?

Answer

In this question, we're told the maximum amount of charge that a capacitor can hold at a given voltage. We're then asked to determine the capacitance. To do this, we'll need to use the expression for capacitance.

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

Plug in the values given to us in the question stem:

$C=\frac{36 C}{12 V}=3F$

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Question

Four arrangements of capacitors are pictured. Each has an equivalent capacitance. Rank these four arrangements from highest equivalent capacitance to lowest. Assume that all capacitors are identical.

Answer

Let's go through all of them and find the equivalent capacitance.

(A) This is just

$C_{eq}=C$

(B) There are two capacitors in series, so this is

$C_{eq}=\left ( \frac{1}{C}+\frac{1}{C}\right )^{-1}=\left ( \frac{2}{C}\right )^{-1}=\frac{1}{2}C$

(C) These capacitors are in parallel, so

$C_{eq}=C+C+C=3C$

(D) These are a combination of series and parallel. Two are in series and they are in parallel with a third,

$C_{eq}=\left ( \frac{1}{C}+\frac{1}{C}\right )^{-1}+C=\left ( \frac{2}{C}\right )^{-1}+C=\frac{3}{2}C$

So, ranking them we get

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Question

Consider the circuit:

What is the total equivalent capacitance?

$C1 = 5F,\ C2 = 7F,\ C3 = 1F$

Answer

Equivalent capacitance in parallel is calculated by taking the sum of each individual capacitor. We can reduce the two parallel capacitors as the following:

$C_{eq}=C2+C3 = 7F+1F= 8F$

The new equivalent circuit has two capacitors in series. This requires us to sum the reciprocals to find equivalent capacitance:

$\frac{1}{C_{tot}}=\sum\frac{1}{C} = \frac{1}{5}+\frac{1}{8} = \frac{13}{40}$

$C_{tot} = 3.08F$

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Question

In this parallel-plate capacitor, the distance between the plates is , the area of each plate is , and the voltage across them is .

Calculate the capacitance.

$\epsilon_0=8.854\cdot 10^{-12}\frac{C^2}{N\cdot m^2}$

Answer

The key hint to remember here is that the capacitance depends only on the geometry of the material, not the potential difference or electric field.

For a parallel-plate capacitor,

$C=\frac{\epsilon _{0}A}{d}$

Plugging in the numbers given results in

$C=\frac{8.854\cdot 10^{-12}\frac{C^2}{N\cdot m^2}\cdot0.0015m^2}{0.03m}$

$C=4.43\cdot 10^{-13}F$

Be careful to convert the units to meters!

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Question

You have 3 capacitors in series. Their capcitance's are $4\mu F$ , $3\mu F$ , and $2\mu F$ . What is the total capacitance of the system?

Answer

To find the total capacitance of capacitors in series, you use the following equation:

$\frac{1}{C_{tot}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...+\frac{1}{C_n}$ .

Our values for , , and are 4, 3, and 2. Now, we can plug in our values to find the answer.

$\frac{1}{C_{tot}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\frac{1}{C_{tot}}=\frac{1}{4}+\frac{1}{3}+\frac{1}{2}$

$\frac{1}{C_{tot}}= \frac{13}{12}$

The answer we have is the inverse of this. Therefore, the total capacitance is $\frac{12}{13} \mu F$ .

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Question

You have 4 capacitors, , , , and , arranged as shown in the diagram below.

Their capacitances are as follows:

$\begin{align} A&=3F \ B&=6F \ C&=4F \ D&=1F \end{align}$

What is the total capacitance of the circuit?

Answer

Remember, the equations for adding capacitances are as follows:

$(Series)\ \frac{1}{C_{tot}}&=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

$(Parallel)\ C_{tot}&=C_1+C_2+...+C_n$

Capacitors and are in series, and are in parallel, and and are in parallel.

$C_{AB}&=(\frac{1}{C_A}+\frac{1}{C_B})^{-1}$

$C_{AB}=(\frac{1}{3}+\frac{1}{6})^{-1}$

$C_{AB}= 2$

$C_{ABC}=C_{AB}+C_C$

$C_{ABC}=2+4$

$C_{ABC}= 6$

$C_{total}=C_{ABC}+C_{D}$

$C_{total}=6+1$

$C_{total}=7$

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Question

Three capacitors are in parallel with each other. What can be said for certain about the total capacitance?

Answer

Capacitors in parallel combine according to the following equation:

$C_{tot}=C_1+C_2+...+C_n$ .

Because the capacitances are additive, and all of the capacitances are greater than zero, no matter what numbers you use, you will always end up with a number that is greater than any of the individual numbers.

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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 voltage difference between the plates increase, decrease or stay the same?

Answer

The fact that the system is still connected to the battery indicate a constant V so regardless what happens to the capacitor, V stays fixed.

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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 charge on each plate increase, decrease or stay the same?

Answer

Relevant equations:

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

Plug the second equation into the first:

$Q=\frac{\varepsilon _{o}AV}{D}$

Considering all the variables in the numerator are held fixed for this problem, we see that increasing D will decrease the charge stored on each plate.

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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 charge on each plate increase, decrease or stay the same?

Answer

The charge has no where to go. Without the battery connected, the charge has no physical avenue on or off the plates.

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

Answer

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

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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Question

Consider the given diagram. If , each plate of the capacitor has surface area , and the plates at 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 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:

$E\epsilon_0A=q$

Convert to and plug in values:

$(4000)(8.8610^{-12})(10010^{-4})=q$

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

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Question

Consider the given diagram. If , each plate of the capacitor has surface area , and the plates are 0.1mm apart, determine the number of excess electrons on the negative plate.

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

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 find the electric field:

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

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:

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

$E\epsilon_0A=q$

Convert to and plug in values:

$(4000)(8.8610^{-12})(10010^{-4})=q$

$3.5410^{-10}=q$

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:

$\frac{3.5410^{-10}}{1.610^{-19}}=n$

$n=2.2110^{9}$

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Question

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 in length and in width, and the space between the plates is .

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 $8.85*10^{-12}\frac{F}{m}$ .

Answer

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

$A=(.5.001)m^2=510^{-4}m^2$

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

Putting it all together,

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

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Question

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 in length and in width, and the space between the plates is .

What is the voltage difference across the capacitor?

Answer

Voltage difference $\Delta V$ is given by

$\Delta V=Ed$ , 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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Question

Consider the circuit:

What is the total equivalent capacitance?

$C1 = 5F,\ C2 = 7F,\ C3 = 1F$

Answer

Equivalent capacitance in parallel is calculated by taking the sum of each individual capacitor. We can reduce the two parallel capacitors as the following:

$C_{eq}=C2+C3 = 7F+1F= 8F$

The new equivalent circuit has two capacitors in series. This requires us to sum the reciprocals to find equivalent capacitance:

$\frac{1}{C_{tot}}=\sum\frac{1}{C} = \frac{1}{5}+\frac{1}{8} = \frac{13}{40}$

$C_{tot} = 3.08F$

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Question

In this parallel-plate capacitor, the distance between the plates is , the area of each plate is , and the voltage across them is .

Calculate the capacitance.

$\epsilon_0=8.854\cdot 10^{-12}\frac{C^2}{N\cdot m^2}$

Answer

The key hint to remember here is that the capacitance depends only on the geometry of the material, not the potential difference or electric field.

For a parallel-plate capacitor,

$C=\frac{\epsilon _{0}A}{d}$

Plugging in the numbers given results in

$C=\frac{8.854\cdot 10^{-12}\frac{C^2}{N\cdot m^2}\cdot0.0015m^2}{0.03m}$

$C=4.43\cdot 10^{-13}F$

Be careful to convert the units to meters!

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Question

You have 3 capacitors in series. Their capcitance's are $4\mu F$ , $3\mu F$ , and $2\mu F$ . What is the total capacitance of the system?

Answer

To find the total capacitance of capacitors in series, you use the following equation:

$\frac{1}{C_{tot}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+...+\frac{1}{C_n}$ .

Our values for , , and are 4, 3, and 2. Now, we can plug in our values to find the answer.

$\frac{1}{C_{tot}} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

$\frac{1}{C_{tot}}=\frac{1}{4}+\frac{1}{3}+\frac{1}{2}$

$\frac{1}{C_{tot}}= \frac{13}{12}$

The answer we have is the inverse of this. Therefore, the total capacitance is $\frac{12}{13} \mu F$ .

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Question

You have 4 capacitors, , , , and , arranged as shown in the diagram below.

Their capacitances are as follows:

$\begin{align} A&=3F \ B&=6F \ C&=4F \ D&=1F \end{align}$

What is the total capacitance of the circuit?

Answer

Remember, the equations for adding capacitances are as follows:

$(Series)\ \frac{1}{C_{tot}}&=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_n}$

$(Parallel)\ C_{tot}&=C_1+C_2+...+C_n$

Capacitors and are in series, and are in parallel, and and are in parallel.

$C_{AB}&=(\frac{1}{C_A}+\frac{1}{C_B})^{-1}$

$C_{AB}=(\frac{1}{3}+\frac{1}{6})^{-1}$

$C_{AB}= 2$

$C_{ABC}=C_{AB}+C_C$

$C_{ABC}=2+4$

$C_{ABC}= 6$

$C_{total}=C_{ABC}+C_{D}$

$C_{total}=6+1$

$C_{total}=7$

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Question

Three capacitors are in parallel with each other. What can be said for certain about the total capacitance?

Answer

Capacitors in parallel combine according to the following equation:

$C_{tot}=C_1+C_2+...+C_n$ .

Because the capacitances are additive, and all of the capacitances are greater than zero, no matter what numbers you use, you will always end up with a number that is greater than any of the individual numbers.

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