AP Physics 2 - Capacitors and Capacitance

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?

$\frac{12}{13} \mu F$

$\frac{13}{12} \mu F$

$9 \mu F$

$\frac{1}{9} \mu F$

None of the other answers is correct

Explanation

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 $\frac{12}{13} \mu F$ .

Lazy capacitor

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.

$3.54*10^{-10}C$

$9.26*10^{-10}C$

$6.11*10^{-10}C$

$1.11*10^{-10}C$

$4.44*10^{-10}C$

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 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:

$E*\epsilon_0*A=q$

Convert to and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

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

Lazy capacitor

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.

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

$n=2.21*10^{9}$

$n=5.85*10^{9}$

$n=5.15*10^{9}$

$n=3.33*10^{9}$

None of these

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 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:

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

$E*\epsilon_0*A=q$

Convert to and plug in values:

$(4000)*(8.86*10^{-12})*(100*10^{-4})=q$

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

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:

$\frac{3.54*10^{-10}}{1.6*10^{-19}}=n$

$n=2.21*10^{9}$

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

4capacitorcircuit

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?

$\frac{36}{49}F$

$\frac{49}{36}F$

Explanation

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.

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

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

$C_{AB}= 2$

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

$C_{ABC}=2+4$

$C_{ABC}= 6$

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

$C_{total}=6+1$

$C_{total}=7$

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

Explanation

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:

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

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

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

$C=2.95*10^{-12}F$

$C=3.38*10^{-6}F$

$C=3.37*10^{6}F$

Explanation

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

$A=(.5*.001)m^2=5*10^{-4}m^2$

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

Putting it all together,

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

Consider the circuit:

Circuit_3__capacitors_

What is the total equivalent capacitance?

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

Explanation

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:

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

$C_{tot} = 3.08F$

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

The total capacitance is greater than any of the individual capacitances.

The total capacitance is less than any of the individual capacitances.

The total capacitance is somewhere between the highest capacitance and the lowest capacitance (inclusive).

Nothing can be said for certain about the total capacitance.

The total capacitance is the arithmetic mean of the individual capacitances.

Explanation

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.

Suppose I have a uniform electric field within a parallel plate capacitor with field strength of $E=41\frac{N}{C}$ .