Question 1

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

Accepted Answer

The capacitance of a capacitor is given by the equation C = Q/V, where Q is the charge stored on each plate and V is the potential difference applied across the plates. In this case, Q = 1 mC = 10^-3 C and V = 100 V. Therefore, C = (10^-3 C)/(100 V) = 10^-5 F or 10 μF (microfarads). The only option that matches this is B, which is 10 μF.

Question 2

The capacitance of a parallel-plate capacitor is:

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 3

Pulling the plates of an isolated charged capacitor apart:

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 4

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:

Accepted Answer

The capacitance of a parallel-plate capacitor is given by:

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

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:

$C = \frac{(8.85 	imes 10^{-12} F/m)(0.2 m^2)}{0.1 	imes 10^{-3} m} = 1.77 	imes 10^{-9} F$

The potential difference across a capacitor is given by:

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

where $Q$ is the magnitude of the charge on each plate.

Substituting the given values, we get:

$V = \frac{(4 	imes 10^{-10} C)}{1.77 	imes 10^{-9} F} = 2.26 V \approx \boxed{	extbf{(D)}\ 2 	imes 10^{12} V}$

Question 5

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

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 6

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

Accepted Answer

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

Question 7

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

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 8

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

Accepted Answer

According to the equation C = εA/d, where C is the capacitance, ε is the permittivity of the medium between the plates, A is the area of the plates, and d is the distance between them, doubling the plate area while keeping the distance constant will result in a doubling of the capacitance. The capacitance is also equal to Q/V, where Q is the charge on each plate and V is the potential difference between them. Therefore, doubling the capacitance while keeping the charge constant will result in halving the potential difference (V).

Question 9

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

Accepted Answer

The capacitance of a parallel-plate capacitor is given by the formula C = ε₀(A/d), where A is the plate area and d is the separation between the plates. If both A and d are doubled, the changes cancel each other out, leaving the capacitance unchanged.

Question 10

The capacitance of a cylindrical capacitor can be increased by:

Accepted Answer

The answer of The capacitance of a cylindrical capacitor can...

Question 11

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

Accepted Answer

The potential difference required to charge a capacitor is given by V = Q/C, where V is the potential difference, Q is the charge, and C is the capacitance. In this case, Q = 2 C and C = 1 F, so V = 2 C / 1 F = 2 V.

Question 12

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

Accepted Answer

The capacitance of a single isolated spherical conductor is given by the formula C = 4πε₀R, where ε₀ is the permittivity of free space. This formula shows that capacitance is directly proportional to the radius R, so the correct answer is A).

Question 13

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:

Accepted Answer

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

Question 14

A farad is the same as a:

Accepted Answer

The farad (F) is a unit of capacitance and is defined as the amount of charge (measured in coulombs, C) that can be stored per unit of voltage (measured in volts, V), so the units are C/V.

Question 15

Two conducting spheres have radii of R₁ and R₂ with R₁ greater than R₂. If they are far apart the capacitance is proportional to:

Accepted Answer

The capacitance between two conducting spheres is given by the formula C = 4πε₀(R1R2)/(R1 + R2), where R1 and R2 are the radii of the spheres and ε₀ is the permittivity of free space. Rearranging, we get C = 4πε₀(R1R2)/(R1 - R2) for the case where the spheres are far apart. Substituting the given values, we get C ∝ (R₁)(R₂)/(R₁ - R₂), which matches option A.

Question 16

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

Accepted Answer

A capacitor stores equal and opposite charges on its two plates, so if one plate has a charge of +Q, the other plate must have a charge of -Q.

Question 17

The capacitance of a spherical capacitor with inner radius a and outer radius b is proportional to:

Accepted Answer

The capacitance of a spherical capacitor is given by the formula $C = 4\pi\epsilon_0 \frac{ab}{b - a}$, where $a$ and $b$ are the inner and outer radii, respectively. This shows that the capacitance is directly proportional to $ab/(b - a)$.

Question 18

The units of capacitance are equivalent to:

Accepted Answer

The answer of The units of capacitance are equivalent to:
...

Question 19

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

Accepted Answer

The capacitance of a parallel-plate capacitor is directly proportional to the plate area and inversely proportional to the plate separation. Therefore, by decreasing the plate separation, the capacitance will increase.

Question 20

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

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.

Quiz 25: Capacitance