Question 1

You want to use three capacitors in a circuit. If each capacitor has a capacitance of 3 pF, the configuration that gives you an equivalent capacitance of 1 pF between points x and y is

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Question 2

A parallel plate capacitor is constructed using two square metal sheets, each of side L = 10 cm. The plates are separated by a distance d = 2 mm and a voltage applied between the plates. The electric field strength within the plates is E = 4000 V/m. The energy stored in the capacitor is

Accepted Answer

The capacitance of a parallel plate capacitor is given by:

$C = \frac{\epsilon_0 A}{d}$

where $\epsilon_0$ is the permittivity of free space, $A$ is the area of each plate, and $d$ is the separation between the plates.

Substituting the given values, we get:

$C = \frac{(8.85 	imes 10^{-12} F/m) \cdot (0.1 m)^2}{0.002 m} = 4.425 	imes 10^{-11} F$

The energy stored in a capacitor is given by:

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

where $V$ is the voltage applied between the plates. Substituting the given values, we get:

$U = \frac{1}{2} (4.425 	imes 10^{-11} F) (V^2)$

We need to find $U$ in joules. Since the SI unit of energy is the joule, we need to use the volt as the unit of voltage. Recall that 1 V = 1 J/C, where J is the joule and C is the coulomb. Therefore, we have:

$U = \frac{1}{2} (4.425 	imes 10^{-11} F) [(V/m)^2] (10^{-2} m)^2$

$U = \frac{1}{2} (4.425 	imes 10^{-11} F) (4000 V/m)^2 (10^{-4} m^2)$

$U = 1.42 	imes 10^{-9} J = 1.42 nJ$

Therefore, the answer is (B), 1.42 nJ.

Question 3

If C₁ < C₂< C₃ < C₄ for the combination of capacitors shown, the equivalent capacitance

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Question 4

You want to use three capacitors in a circuit. If each capacitor has a capacitance of 2 pF, the configuration that gives you an equivalent capacitance of 3 pF between points x and y is

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The answer of  You want to use three capacitors...

Question 5

The equivalent capacitance of two capacitors in series is

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The equivalent capacitance of two capacitors in series is always less than the smaller of the capacitances because the total capacitance decreases as more capacitors are added in series.

Question 6

The voltage across each capacitor in a set of capacitors in parallel is

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When capacitors are connected in parallel, the voltage across each capacitor is the same. This is because the potential difference between the two points where the capacitors are connected is the same, and since the capacitors are connected in parallel, they have the same voltage drop. Therefore, the voltage across each capacitor is the same and not dependent on its capacitance.

Question 7

The equivalent capacitance of three capacitors in series is

Accepted Answer

When capacitors are connected in series, the equivalent capacitance is always less than the smallest individual capacitor in the series. This is because the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.

Question 8

You connect a 6-μF capacitor in parallel to two 12-μF capacitors that are connected in series. The equivalent capacitance of the combination is

Accepted Answer

The two 12-μF capacitors in series have an equivalent capacitance of 6 μF (1/C_eq = 1/12 + 1/12). This 6-μF is in parallel with the 6-μF capacitor, giving a total equivalent capacitance of 12 μF (6 + 6).

Question 9

You connect two 12-μF capacitors and a 6-μF capacitor in parallel. The equivalent capacitance of the combination is

Accepted Answer

When capacitors are connected in parallel, their capacitances add up. Therefore, the equivalent capacitance is 12 μF + 12 μF + 6 μF = 30 μF.

Question 10

The charge on each capacitor in a set of capacitors in series is

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In a series circuit, the charge on each capacitor is the same because the same amount of charge flows through each component in the series without any branching.

Question 11

A 1.0-μF capacitor and a 2.0-μF capacitor are connected in series across a 1200-V source. The charge on each capacitor is

Accepted Answer

The voltage across each capacitor will be different as they have different capacitances. Let the voltage across the 1.0 μF capacitor be V1 and the voltage across the 2.0 μF capacitor be V2. We know that the total voltage across the two capacitors is 1200 V, so we can write the equation:
V1 + V2 = 1200
We also know that the charge on each capacitor is given by Q = CV, where C is the capacitance and V is the voltage across the capacitor. So the charge on the 1.0 μF capacitor is Q1 = (1.0 μF)(V1) and the charge on the 2.0 μF capacitor is Q2 = (2.0 μF)(V2). 
Since the capacitors are in series, the charge on each capacitor must be the same. So we can write:
Q1 = Q2
Substituting Q = CV into this equation and rearranging, we get:
C1V1 = C2V2
We can solve these two equations for V1 and V2 to get: 
V1 = 800 V and V2 = 400 V
The charges on each capacitor are therefore:
Q1 = (1.0 μF)(800 V) = 0.80 mC
Q2 = (2.0 μF)(400 V) = 0.80 mC
So the answer is B.

Question 12

You connect two 12-μF capacitors and a 6-μF capacitor in series. The equivalent capacitance of the combination is

Accepted Answer

When capacitors are connected in series, the equivalent capacitance is found using the formula:

1/Ceq = 1/C1 + 1/C2 + ...

In this case, we have three capacitors in series:

1/Ceq = 1/12 μF + 1/12 μF + 1/6 μF

Simplifying this expression gives:

1/Ceq = 1/6 μF

Therefore, the equivalent capacitance is:

Ceq = 6 μF.

Question 13

You want to use three capacitors in a circuit. If each capacitor has a capacitance of 3 pF, the configuration that gives you an equivalent capacitance of 2 pF between points x and y is

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Question 14

You connect three capacitors as shown in the diagram below. The effective capacitance of this combination when C₁ = 5.0 μF, C₂ = 4.0 μF, and C₃ = 3.0 μF is approximately

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Question 15

The voltage across each capacitor in a set of capacitors in series is

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In a set of capacitors in series, the voltage across each capacitor depends on its capacitance and the total voltage across the set of capacitors. The voltage is inversely proportional to the capacitance, meaning that the larger the value of capacitance, the lower the voltage across that capacitor. This is because the total charge on the capacitors in series is the same, and as capacitance increases, the capacitance has more ability to store charge, which lowers the voltage.

Question 16

If C₁ < C₂ < C₃ < C₄ for the combination of capacitors shown, the equivalent capacitance

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Question 17

The charge on each capacitor in a set of capacitors in parallel is

Accepted Answer

In a parallel circuit, the voltage across each capacitor is the same. Since charge (Q) is equal to capacitance (C) times voltage (V) (Q = CV), the charge on each capacitor is directly proportional to its capacitance.

Question 18

You connect three capacitors as shown in the diagram below. If the potential difference between A and B is 24.5-V, what is the total energy stored in this system of capacitors if C₁ = 5.0 μF, C₂ = 4.0 μF, and C₃ = 3.0 μF?

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The total energy stored in the system of capacitors can be found using the formula:
$E=\frac{1}{2}\left(\frac{Q_1^2}{C_1}+\frac{Q_2^2}{C_2}+\frac{Q_3^2}{C_3}ight)$
where $Q_1$, $Q_2$, and $Q_3$ are the charges stored on the respective capacitors, and $C_1$, $C_2$, and $C_3$ are their capacitances. 
The charges on all capacitors are equal and can be found using:
$Q=CV$
where $V$ is the potential difference between A and B. 
$Q=CV=5.0	imes10^{-6}	ext{ F}	imes24.5	ext{ V}=1.225	imes10^{-4}	ext{ C}$
Now we can calculate the energy:
$E=\frac{1}{2}\left(\frac{(1.225	imes10^{-4}	ext{ C})^2}{5.0	imes10^{-6}	ext{ F}}+\frac{(1.225	imes10^{-4}	ext{ C})^2}{4.0	imes10^{-6}	ext{ F}}+\frac{(1.225	imes10^{-4}	ext{ C})^2}{3.0	imes10^{-6}	ext{ F}}ight)$
$E=6.8	imes10^{-4}	ext{ J}=6.8	imes10^{-4}	imes10^{-10}	ext{PJ}=6.8	imes10^{-14}	ext{ PJ}$
Therefore, the correct answer is option D.

Question 19

You connect two capacitors C₁ = 15 pF and C₂ = 30 pF in series across a 1.5-V battery. The potential difference across capacitor C₁ is approximately

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Question 20

You connect three capacitors as shown in the diagram. C₁ = 5.0 μF, C₂ = 4.0 μF, and C₃ = 3.0 μF. If you apply 12-V between points A and B, the energy stored in C₃ will be approximately

Accepted Answer

The energy stored in a capacitor is given by the formula: E = 1/2 * C * V^2 where E is the energy, C is the capacitance, and V is the voltage. In this circuit, the capacitors are in series, so the total capacitance is: Ct = 1/(1/C1 + 1/C2 + 1/C3) Plugging in the values, we get: Ct = 1/(1/5 + 1/4 + 1/3) = 60/31 μF The voltage across the capacitors is 12V, since they are in series. Now we can calculate the energy stored in C₃: E = 1/2 * 3.0 μF * (12V)^2 / (60/31 μF) = 0.118 mJ Therefore, the closest answer is C - 0.12 mJ.

Quiz 17: Electrostatics II: Electric Potential Energy and Electric Potential