Question 1

A farad is the same as a:
A)J/V
B)V/J
C)C/V
D)V/C
E)N/C

Accepted Answer

A farad is the unit of capacitance, which is defined as the ratio of electric charge (measured in coulombs) to electric potential difference (measured in volts). Therefore, the unit of farad can be expressed as C/V (charge per unit voltage).

Question 2

If the plate area of an isolated charged parallel-plate capacitor is doubled:
A)the electric field is doubled
B)the potential difference is halved
C)the charge on each plate is halved
D)the surface charge density on each plate is doubled
E)none of the above

Accepted Answer

B

Question 3

The units of capacitance are equivalent to: A)J/C B)V/C C)J²/C D)C/J E)C²/J

Accepted Answer

E

Question 4

A parallel-plate capacitor C has a charge Q.The actual charges on its plates are:
A)Q, Q
B)Q/2, Q/2
C)Q, -Q
D)Q/2, -Q/2
E)Q, 0

Accepted Answer

In a parallel-plate capacitor, the charges on the two plates are equal in magnitude but opposite in sign, hence one plate has a charge of +Q and the other has a charge of -Q.

Question 5

The capacitance of a cylindrical capacitor can be increased by:
A)decreasing both the radius of the inner cylinder and the length
B)increasing both the radius of the inner cylinder and the length
C)increasing the radius of the outer cylindrical shell and decreasing the length
D)decreasing the radius of the inner cylinder and increasing the radius of the outer cylindrical shell
E)only by decreasing the length

Accepted Answer

The capacitance of a cylindrical capacitor is directly proportional to the length and the radius of the inner cylinder, and inversely proportional to the distance between the inner and outer cylinders. Therefore, increasing both the radius of the inner cylinder and the length will increase the capacitance.

Question 6

To charge a 1-F capacitor with 2 C requires a potential difference of:
A)0.2 V
B)0.5 V
C)2 V
D)5 V
E)none of these

Accepted Answer

We can use the formula Q=CV, where Q is the charge stored in the capacitor, C is the capacitance, and V is the potential difference across the capacitor.

In this case, Q = 2 C and C = 1 F, so:

2 C = (1 F)(V)
V = 2 C/1 F
V = 2 V

Therefore, the potential difference required to charge the 1-F capacitor with 2 C is 2 V, which corresponds to choice C.

Question 7

The capacitance of a parallel-plate capacitor can be increased by:
A)increasing the charge
B)decreasing the charge
C)increasing the plate separation
D)decreasing the plate separation
E)decreasing the plate area

Accepted Answer

The capacitance of a parallel-plate capacitor is directly proportional to the plate area and inversely proportional to the distance between the plates (plate separation). Therefore, decreasing the plate separation will increase the capacitance.

Question 8

Pulling the plates of an isolated charged capacitor apart:
A)increases the capacitance
B)increases the potential difference
C)does not affect the potential difference
D)decreases the potential difference
E)does not affect the capacitance

Accepted Answer

Pulling the plates of an isolated charged capacitor apart increases the distance between the plates without changing the charge on them. According to the formula $C = \frac{Q}{V}$, where $C$ is the capacitance, $Q$ is the charge, and $V$ is the potential difference, keeping $Q$ constant while increasing the distance between the plates (which decreases $C$) must increase $V$, the potential difference.

Question 9

A parallel-plate capacitor has a plate area of 0.2 m² and a plate separation of 0.1 mm.To obtain an electric field of 2.0 *10⁶ V/m between the plates, the magnitude of the charge on each plate should be: A)3.5 * 10^-6C B)7.1 *10^-6C C)1.4 * 10^-5C D)1.8 * 10^-5C E)8.9 * 10^-5C

Accepted Answer

The answer of A parallel-plate capacitor has a plate area...

Question 10

A parallel-plate capacitor has a plate area of 0.3 m² and a plate separation of 0.1 mm.If the charge on each plate has a magnitude of 5 *10^-6 C then the force exerted by one plate on the other has a magnitude of about: A)5 N B)9 N C)1 * 10⁴ N D)9 *10⁵ N E)2* 10⁷ N

Accepted Answer

The answer of A parallel-plate capacitor has a plate area...

Question 11

If both the plate area and the plate separation of a parallel-plate capacitor are doubled, the capacitance is:
A)doubled
B)halved
C)unchanged
D)one-fourth the original
E)quadrupled

Accepted Answer

The capacitance of a parallel-plate capacitor is given by the equation C = εA/d, where ε is the permittivity of the material between the plates, A is the area of each plate, and d is the separation distance between the plates. If we double both A and d, we get a new capacitance of C' = ε(2A)/(2d) = (εA/d) = C, which is equal to the original capacitance. Therefore, the capacitance is unchanged.

Question 12

Each plate of a capacitor stores a charge of magnitude 1 mC when a 100-V potential difference is applied.The capacitance is:
A)5 $\mu$F
B)10 $\mu$F
C)50 $\mu$F
D)100 $\mu$F
E)none of these

Accepted Answer

The charge stored on a capacitor is given by Q = CV, where Q is the charge, C is the capacitance and V is the potential difference. Here, we are given Q = 1 mC = 10^-3 C and V = 100 V. Thus, C = Q/V = (10^-3)/100 = 10^-5 F = 10 μF. The only option that has 10 μF as an answer is option B.

Question 13

Each of the four capacitors shown is 500 $\mu$F.The voltmeter reads 1000V.The magnitude of the charge on each capacitor plate is: 
A)0.2 C
B)0.5 C
C)20 C
D)50 C
E)none of these

Accepted Answer

The answer of Each of the four capacitors shown is...

Question 14

If the plate separation of an isolated charged parallel-plate capacitor is doubled:
A)the electric field is doubled
B)the potential difference is halved
C)the charge on each plate is halved
D)the surface charge density on each plate is doubled
E)none of the above

Accepted Answer

Doubling the plate separation of an isolated charged parallel-plate capacitor does not double the electric field, halve the potential difference, halve the charge on each plate, or double the surface charge density. Instead, it doubles the potential difference (V) because the electric field (E) in a parallel-plate capacitor is given by E = V/d, where d is the separation between the plates. For an isolated capacitor, the charge (Q) remains constant, and since the capacitance (C) is given by C = ε₀A/d (where A is the area of the plates and ε₀ is the permittivity of free space), doubling d halves the capacitance. Since Q = CV, if C is halved and Q is constant, V must double.

Question 15

A parallel-plate capacitor has a plate area of 0.2 m² and a plate separation of 0.1 mm.If the charge on each plate has a magnitude of 4 *10^-6 C the potential difference across the plates is approximately: A)0 V B)4 * 10^-2 V C)2 * 10² V D)2 *10⁵ V E)4 * 10⁸ V

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:

$C = \frac{(8.85	imes10^{-12}	ext{ F/m}) (0.2	ext{ m}^2)}{0.1	imes10^{-3}	ext{ m}} = 1.77	imes10^{-8}	ext{ F}$

The potential difference across the plates is given by:

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

where $Q$ is the charge on each plate. Substituting the given values, we get:

$V = \frac{4	imes10^{-6}	ext{ C}}{1.77	imes10^{-8}	ext{ F}} = 226.757	ext{ V} \approx 2.2	imes10^2	ext{ V} \approx 2	ext{ V}$

Therefore, the best choice is (C), which is closest to the calculated value of 2 V.

Question 16

The plate areas and plate separations of five parallel plate capacitors are  Rank these according to their capacitances, least to greatest.
A)1, 2, 3, 4, 5
B)5, 4, 3, 2, 1
C)5, then 3 and 4 tie, then 1, then 2
D)4, then 1 and 2 tie, then 5, then 3
E)3, then 5, then 1 and 2 tie, then 4

Accepted Answer

The answer of The plate areas and plate separations of...

Question 17

The capacitance of a parallel-plate capacitor is:
A)proportional to the plate area
B)proportional to the charge stored
C)independent of any material inserted between the plates
D)proportional to the potential difference of the plates
E)proportional to the plate separation

Accepted Answer

The capacitance of a parallel-plate capacitor is directly proportional to the plate area. This is because the capacitance is defined as the ratio of the charge stored on the plates to the potential difference between them, and the charge stored on the plates is proportional to the plate area.

Question 18

If the charge on a parallel-plate capacitor is doubled:
A)the capacitance is halved
B)the capacitance is doubled
C)the electric field is halved
D)the electric field is doubled
E)the surface charge density is not changed on either plate

Accepted Answer

Doubling the charge on a parallel-plate capacitor directly doubles the surface charge density, which in turn doubles the electric field between the plates, as the electric field (E) is directly proportional to the surface charge density (σ) according to the equation E = σ / ε₀, where ε₀ is the permittivity of free space. Capacitance (C) is defined as the charge (Q) divided by the voltage (V) and does not change with the amount of charge alone, as it is a property of the geometry and the dielectric material of the capacitor.

Question 19

The capacitance of a parallel-plate capacitor with plate area A and plate separation d is given by: A)"$\varepsilon$₀d/A" B)"$\varepsilon$₀d/2A" C)"$\varepsilon$₀A/d" D)"$\varepsilon$₀A/2d" E)"$\varepsilon$d/$\varepsilon$₀"

Accepted Answer

The capacitance of a parallel-plate capacitor is given by the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of one of the plates, and d is the separation between the plates.

Question 20

The capacitance of a single isolated spherical conductor with radius R is proportional to: A)R B)R² C)1/R D)1/R² E)none of these

Accepted Answer

The capacitance of a spherical conductor is given by C = 4πεR, where ε is the permittivity of free space. Therefore, the capacitance is directly proportional to the radius R.

Quiz 25: Capacitance

A farad is the same as a:

If the plate area of an isolated charged parallel-plate capacitor is doubled:

The units of capacitance are equivalent to:

A parallel-plate capacitor C has a charge Q.The actual charges on its plates are:

The capacitance of a cylindrical capacitor can be increased by:

To charge a 1-F capacitor with 2 C requires a potential difference of:

The capacitance of a parallel-plate capacitor can be increased by:

Pulling the plates of an isolated charged capacitor apart:

A parallel-plate capacitor has a plate area of 0.2 m2 and a plate separation of 0.1 mm.To obtain an electric field of 2.0 *106 V/m between the plates, the magnitude of the charge on each plate should be:

A parallel-plate capacitor has a plate area of 0.3 m2 and a plate separation of 0.1 mm.If the charge on each plate has a magnitude of 5 *10-6 C then the force exerted by one plate on the other has a magnitude of about:

If both the plate area and the plate separation of a parallel-plate capacitor are doubled, the capacitance is:

Each plate of a capacitor stores a charge of magnitude 1 mC when a 100-V potential difference is applied.The capacitance is:

Each of the four capacitors shown is 500 μ\muμ F.The voltmeter reads 1000V.The magnitude of the charge on each capacitor plate is:

If the plate separation of an isolated charged parallel-plate capacitor is doubled:

A parallel-plate capacitor has a plate area of 0.2 m2 and a plate separation of 0.1 mm.If the charge on each plate has a magnitude of 4 *10-6 C the potential difference across the plates is approximately:

The plate areas and plate separations of five parallel plate capacitors are Rank these according to their capacitances, least to greatest.

The capacitance of a parallel-plate capacitor is:

If the charge on a parallel-plate capacitor is doubled:

The capacitance of a parallel-plate capacitor with plate area A and plate separation d is given by:

The capacitance of a single isolated spherical conductor with radius R is proportional to:

A parallel-plate capacitor has a plate area of 0.2 m² and a plate separation of 0.1 mm.To obtain an electric field of 2.0 *10⁶ V/m between the plates, the magnitude of the charge on each plate should be:

A parallel-plate capacitor has a plate area of 0.3 m² and a plate separation of 0.1 mm.If the charge on each plate has a magnitude of 5 *10^-6 C then the force exerted by one plate on the other has a magnitude of about:

A parallel-plate capacitor has a plate area of 0.2 m² and a plate separation of 0.1 mm.If the charge on each plate has a magnitude of 4 *10^-6 C the potential difference across the plates is approximately: