Electric Circuits

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AP Physics C: Electricity and Magnetism › Electric Circuits

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If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

$C_{eq}=\frac{C}{2}$

$C_{eq}=C$

$C_{eq}= 2C$

$C < C_{eq}< 2C$

$\frac{C}{2} < C_{eq} < C$

Explanation

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

$C_{eq}=\frac{C}{2}$

$C_{eq}=C$

$C_{eq}= 2C$

$C < C_{eq}< 2C$

$\frac{C}{2} < C_{eq} < C$

Explanation

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

If two identical parallel plate capacitors of capacitance are connected in series, which is true of the equivalent capacitance, $C_{eq}$ ?

$C_{eq}=\frac{C}{2}$

$C_{eq}=C$

$C_{eq}= 2C$

$C < C_{eq}< 2C$

$\frac{C}{2} < C_{eq} < C$

Explanation

Relevant equations:

$\frac{1}{C_{eq, series}}=\frac{1}{C_1}+\frac{1}{C_2}+ ...$

Use the series equation, replacing C1 and C2 with the given constant C:

$\frac{1}{C_{eq, series}}= \frac{1}{C}+ \frac{1}{C}$

$\frac{1}{C_{eq, series}}=\frac{2}{C}$

$C_{eq, series} = \frac{C}{2}$

This agrees with the general rule that the equivalent capacitance in series is less than the capacitance of any of the individual capacitors.

Two parallel conducting plates each have an area of and are separated by a distance, . One plate has a charge evenly distributed across it, and the other has a charge of . A proton (charge ) is initially held near the positive plate, then released such that it accelerates towards the negative plate. How much kinetic energy has the proton gained in this process?

$\frac{eQd}{\epsilon _oA}$

$\frac{e \epsilon _o A}{Qd}$

$\frac{eQ}{\epsilon _o A d}$

$\frac{eQ\epsilon _o A}{d}$

$\frac{eQ \epsilon _o A}{d^2}$

Explanation

Relevant equations:

$W = q\Delta V$

$W = \Delta KE$

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

$C = \frac{\epsilon _o A}{d}$

According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.

1. Find an expression for potential difference, , in terms of and , by setting the two capacitance equations equal to each other:

$\frac{Q}{V}=\frac{\epsilon _o A}{d}$

$V = \frac{Qd}{\epsilon _o A}$

2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):

$W = \frac {eQd}{\epsilon _o A} = \Delta K$

$\frac{eQd}{\epsilon _oA}$

$\frac{e \epsilon _o A}{Qd}$

$\frac{eQ}{\epsilon _o A d}$

$\frac{eQ\epsilon _o A}{d}$

$\frac{eQ \epsilon _o A}{d^2}$

Explanation

Relevant equations:

$W = q\Delta V$

$W = \Delta KE$

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

$C = \frac{\epsilon _o A}{d}$

According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.

1. Find an expression for potential difference, , in terms of and , by setting the two capacitance equations equal to each other:

$\frac{Q}{V}=\frac{\epsilon _o A}{d}$

$V = \frac{Qd}{\epsilon _o A}$

2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):

$W = \frac {eQd}{\epsilon _o A} = \Delta K$

$\frac{eQd}{\epsilon _oA}$

$\frac{e \epsilon _o A}{Qd}$

$\frac{eQ}{\epsilon _o A d}$

$\frac{eQ\epsilon _o A}{d}$

$\frac{eQ \epsilon _o A}{d^2}$

Explanation

Relevant equations:

$W = q\Delta V$

$W = \Delta KE$

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

$C = \frac{\epsilon _o A}{d}$

According to the work-energy theorem, work done on the proton is equal to its change in kinetic energy.

1. Find an expression for potential difference, , in terms of and , by setting the two capacitance equations equal to each other:

$\frac{Q}{V}=\frac{\epsilon _o A}{d}$

$V = \frac{Qd}{\epsilon _o A}$

2. Multiply the potential difference times the proton charge to find work (and thus kinetic energy):

$W = \frac {eQd}{\epsilon _o A} = \Delta K$

Charge is distributed uniformly over the area of the two plates of a parallel plate capacitor, resulting in a surface area charge density of $\pm \sigma$ on the plates (the top plate is positive and the bottom is negative, as shown below). Each plate has area and are separated by distance . A material of dielectric constant $\kappa_{material}$ has been placed between the two plates.

Ps0_capacitor

Which of the following would not result in an increase in the measure of electric potential difference between the two plates?

Increase the area, , of the plates

Increase the value of the surface charge density, $\sigma$ , on each plate

Increase the distance, , between the plates

Replace the material between the two plates with one of a lower dielectric constant, $\kappa$

Explanation

The electric potential difference created between the plates of a parallel plate capactor is given by the equation:

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

The charge can be calculated by using the equation:

$Q=\sigma A$

The value of the capacitance is related to the dimensions of the capacitor with the equation:

$C=\kappa \frac{\varepsilon _{o}A}{d}$

Combining these equations yields:

$V=\frac{\sigma A}{\frac{\kappa \epsilon_0A}{d}}=\frac{\sigma d}{\varepsilon_{o} \kappa }$

The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.

A parallel plate capacitor has a capacitance of . If the plates are apart, what is the area of the plates?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

$1.1*10^{11}m^2$

$1.1*10^{10}m^2$

Explanation

The relationship between capacitance, distance, and area is $C=\epsilon_0 \frac{A}{d}$ . We can rearrange this equation to solve for area.

$A=\frac{Cd}{\epsilon_0}$

Now, we can use the values given in teh question to solve.

$A=\frac{1F(1*10^{-2}m)}{8.85* 10^{-12} \frac{F}{m}}=1.1*10^9m^2$

A parallel plate capacitor has a capacitance of . If the plates are apart, what is the area of the plates?

$\epsilon_0=8.85* 10^{-12} \frac{F}{m}$

$1.1*10^{11}m^2$

$1.1*10^{10}m^2$

Explanation

The relationship between capacitance, distance, and area is $C=\epsilon_0 \frac{A}{d}$ . We can rearrange this equation to solve for area.

$A=\frac{Cd}{\epsilon_0}$

Now, we can use the values given in teh question to solve.

$A=\frac{1F(1*10^{-2}m)}{8.85* 10^{-12} \frac{F}{m}}=1.1*10^9m^2$

Ps0_capacitor

Which of the following would not result in an increase in the measure of electric potential difference between the two plates?

Increase the area, , of the plates

Increase the value of the surface charge density, $\sigma$ , on each plate

Increase the distance, , between the plates

Replace the material between the two plates with one of a lower dielectric constant, $\kappa$

Explanation

The electric potential difference created between the plates of a parallel plate capactor is given by the equation:

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

The charge can be calculated by using the equation:

$Q=\sigma A$

The value of the capacitance is related to the dimensions of the capacitor with the equation:

$C=\kappa \frac{\varepsilon _{o}A}{d}$

Combining these equations yields:

$V=\frac{\sigma A}{\frac{\kappa \epsilon_0A}{d}}=\frac{\sigma d}{\varepsilon_{o} \kappa }$

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