Circuit Components - AP Physics 2

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Question

Fill in the blanks.

Dielectrics capacitance because capacitance is related to electric field strength, and dielectrics effective electric field strength.

Answer

The reason dielectrics reduce the effective electric field strength is because when a dielectric is added, the medium gets polarized, which produces an electric field in opposition of the existing electric field.

When the electric field decreases, the capacitance increases because the plates are able to store more charge for the same amount of voltage applied, because there's less force repelling electrons from gathering on the plate.

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Question

Two parallel conducting plates separated by a distance are connected to a battery with voltage . If the distance between them is doubled and the battery stays connected, which of the following statements are correct?

Answer

The equation for capacitance of parallel conducting plates is:

$C=\frac{\epsilon_0A}{d}$

When the distance is doubled, the capacitance changes to:

$C'=\frac{\epsilon_0A}{d'}=\frac{\epsilon_0A}{2d}=\frac{1}{2}C$

The battery is still connected to the plates, the potential difference is unchanged. Because , and the capacitance is halved while the voltage stays the same, must necessarily drop to half to account for the change.

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Question

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

Answer

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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Question

Capacitances are as follows:

$\begin{align} A&=3F \ B&=7F \ C&=4F \ D&=1F \end{align}$

What is the total capacitance of the system in the diagram above?

Answer

Recall the equations used for adding capacitors:

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

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

From the diagram, capacitors A and B are in parallel, and capacitors C and D are in parallel, and those two systems are in series.

Use the equation above to find the equivalent capacitance of capacitors A and B.

$C_{AB}&=C_A+C_B$

$C_{AB}= 3F+7F$

$C_{AB}= 10F$

Use the equation above to find the equivalent capacitance of capacitors C and D.

$C_{CD}&=C_C+C_D$

$C_{CD}=4F+1F$

$C_{CD}= 5F$

Use the equation above to find the total capacitance by adding the two systems of capacitors, which are in series.

$C_{ABCD}=\left(\frac{1}{C_{AB}}+\frac{1}{C_{CD}}\right)^{-1}$

$C_{ABCD}=\left(\frac{1}{10F}+\frac{1}{5F}\right)^{-1}$

$C_{ABCD}= \frac{10}{3}F$

Therefore, the total capacitance is $\frac{10}{3} F$

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Question

Capacitances are as follows:

$\begin{align} A&=1F \ B&=3F \ C&=4F \ D&=2F \end{align}$

Consider the diagram above. If the battery has a potential difference of , what is the total energy of the system once it's fully charged?

Answer

The equations for adding capacitors are:

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

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

To find the total energy, we need to know the total capacitance. To do that, we the capacitors together according to the rules above.

Capacitors A and B are in parallel.

$C_{AB}=C_A+C_B$

$C_{AB}= 1F+3F$

$C_{AB}= 4F$

Capacitors C and D are in parallel.

$C_{CD}&=C_C+C_D$

$C_{CD}=4F+2F$

$C_{CD}= 6F$

Add the systems of capacitors together. They are in series.

$C_{ABCD}=\left(\frac{1}{C_{AB}}+\frac{1}{C_{CD}}\right)^{-1}$

$C_{ABCD}=\left(\frac{1}{4F}+\frac{1}{6F}\right)^{-1}$

$C_{ABCD}= \frac{12}{5}F$

The total capacitance is $\frac{12}{5}F$

The equation for finding the energy of a capacitor is:

$U=\frac{1}{2}CV^2$

Plug in known values and solve.

$U=\frac{1}{2}CV^2$

$U=\frac{1}{2}\left(\frac{12}{5}F\right)\left((5V)^2\right)$

Therefore, the system, when fully charged, holds of energy.

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Question

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

Suppose the capacitor's plates are in length and in width, and the space between the plates is .

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?

Answer

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 $\Delta V$ ,

$\Delta V=\frac{Q}{C}=\frac{10^{-6}C}{2.9510^{-13}F}=3.4*10^{6}V$

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Question

Consider the circuit:

If the voltage drop across C2 is 5V, what is the total energy stored in C2 and C3?

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

Answer

In parallel branches of a circuit, the voltage drops are all the same. Therefore, we know that the voltage drop across C3 is also 5V.

We can then use the following equation to calculate the total stored energy:

$U=\frac{1}{2}CV^2$

Since the voltage is the same for both capacitors, we can simply add the two capacitances to do one calculation for energy:

$U = \frac{1}{2}(8F)(5V)^2 = 100J$

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Question

If , and , how much energy is stored in ?

Answer

In this circuit, the voltage source, and , and are all in parallel, meaning they share the same voltage.

To find the energy, we can use the formula

$U=\frac{1}2CV^2$ , with being the energy, being the capacitance, and being the voltage drop across that capacitor.

To use the formula we need the voltage across .

Another hint we can use is that and having the same charge since they're in series. First let's find the equivalent capacitance:

$C_{12}= \frac{C_{1}C_{2}}{C_{1}+C_{2}}=\frac{4}{5}F$

Now, we can use the formula

to calculate charge in the capacitor.

$Q=CV=(\frac{4}{5})(5)=4 C$

Now that we know a charge of exists in both capacitors, we can use the formula again to find the voltage in only .

$Q_{1}=C_{1}\cdot V_{1}$

$V_{1}=\frac{4}{4}=1V$

Finally, we plug this into the first equation to calculate energy.

$U=\frac{1}{2}CV^2=\frac{1}{2}(4)(1)^2=2J$

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Question

A $15\mu F$ capacitor is connected to a battery. Once the capacitor is fully charged, how much energy is stored?

Answer

To find the amount of energy stored in a capactior, we use the equation

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

We're given the capacitance (), and the voltage (), so we'll use the third equation.

$U=\frac{1}{2}(15\mu F)(9V)$

$U= 6.075\cdot 10^{-4}J$

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Question

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

Their capacitances are as follows:

$\begin{align} A&=1F \ B&=3F \ C&=2F \ D&=4F \end{align}$

If you have a 6V battery connected to the circuit, what's the total energy stored in the capacitors?

Answer

The equation for energy stored in a capacitor is

$W=\frac{Q^2}{2C}=\frac{CV^2}{2}=\frac{QV}{2}$

We can find the capacitance by adding the capacitors together, and we have the voltage, so we'll use the second equation, $W=\frac{1}{2}CV^2$ .

When adding capacitors, remember how to add in series and parallel.

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

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

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

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

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

$C_{AB}= \frac{3}{4}$

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

$C_{ABC}=\frac{3}{4}+2$

$C_{ABC}= \frac{11}{4}$

$C_{tot}&=C_{ABC}+C_{D}$

$C_{total}=\frac{11}{4}+4$

$C_{total}=\frac{27}{4}$

Now that we have the total capacitance, we can use the earlier equation to find the energy.

$W=\frac{1}{2}CV^2$

$W= \frac{1}{2}(\frac{27}{4})(6)^2$

The total energy stored is 121.5J.

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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. If some external agent pulls the capacitor apart such that D doubles, did the internal energy, U, stored in the capacitor increase, decrease or stay the same?

Answer

Equations required:

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

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

We see from the first equation the when D is doubled, C will be halved (since and $\varepsilon_o$ are constant). From the third equation, we seed that when C is halved, the potential energy, U will double.

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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 internal energy, U, stored in the capacitor increase, decrease or stay the same?

Answer

Relevant equations:

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

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

We must note for this problem that the voltage, V, is kept constant by the battery. Looking at the second equation, if C changes (by changing D) then Q must change when V is held constant. This means that our formula for U must be altered. How can we make claims about U if Q and C are both changing? We need a constant variable in the equation for U so that we can make a direct relationship between D and U. If we plug in the second equation into the third we arrive at:

$U=\frac{1}{2}CV^{2}$

Now, we know that V is held constant by the battery, so when C decreases (because D doubled) we see that U actually decreases here. So it matters whether or not the capacitor is hooked up to a battery.

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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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