Question 1

The potential difference between the plates of a parallel plate capacitor with the plate separation of 6 cm is 60 V. What is the electric field between the plates of this capacitor?

Accepted Answer

The formula for the electric field between the plates of a parallel plate capacitor is given by:

E = V/d

where E is the electric field, V is the potential difference, and d is the plate separation.

Substituting the given values, we get:

E = 60 V/0.06 m = 1000 V/m

Therefore, the electric field between the plates of this capacitor is 1000 V/m, which corresponds to choice C.

Question 2

A parallel plate capacitor has a potential difference between the plates of 80 V. If the charge on one of the plates of the capacitor is +8.0 μC, what is the electrical energy stored by this capacitor?

Accepted Answer

The energy stored in a capacitor is given by the formula U = (1/2)QV, where Q is the charge and V is the voltage. Substituting the given values, U = (1/2)(8.0 × 10^-6 C)(80 V) = 320 × 10^-6 J.

Question 3

FIGURE 20-12 
-Two point charges of magnitude +4.0 μC and -4.0 μC are placed as shown in Figure 20-12. What is the potential difference between points A and B?

Accepted Answer

The answer of FIGURE 20-12 
-Two point charges of magnitude...

Question 4

The potential difference between the plates of a parallel plate capacitor is 35 V and the electric field between the plates has a strength of 750 V/m. If the plate area is 4.0 × 10^-2 m², what is the capacitance of this capacitor?

Accepted Answer

The capacitance of a parallel plate capacitor is given by:

C = εA/d

where ε is the permittivity of the dielectric material between the plates, A is the area of each plate, and d is the distance between the plates.

We are given the potential difference V = 35 V and the electric field strength E = 750 V/m. We can use these values to find the distance between the plates:

V = Ed
35 = (750)(d)
d = 0.0467 m

Now we can calculate the capacitance:

C = (8.85×10^-12)(4.0×10^-2) / 0.0467
C = 7.6×10^-12 F

Therefore, the answer is B.

Question 5

FIGURE 20-14 
-The equipotential surfaces between two spherical conductors are shown in Figure 20-14, with the value of the potential marked for each line. What is the potential difference between points G and D?

Accepted Answer

The potential difference between two points is given by the difference in their potentials.

Potential at point G = +80 V 
Potential at point D = -80 V

Therefore, potential difference between points G and D = -80 V - (+80 V) = -160 V.

So, the correct answer is B.

Question 6

A 20-μF capacitor has a charge of 60 μC. How much energy is stored in this capacitor?

Accepted Answer

The formula for calculating the energy stored in a capacitor is:

E = (1/2) CV^2

Where E is the energy, C is the capacitance, and V is the voltage across the capacitor.

To find the voltage, we can use the formula:

Q = CV

Where Q is the charge on the capacitor.

V = Q/C = 60 μC / 20 μF = 3 V

Now we can use the energy formula to calculate the energy stored in the capacitor:

E = (1/2) CV^2 = (1/2) * 20 μF * (3 V)^2 = 90 μJ

Therefore, the answer is C.

Question 7

What charge appears on the plates of a 2.0-μF capacitor when it is charged to 100 V?

Accepted Answer

The charge (Q) on a capacitor is given by the formula Q = C × V, where C is the capacitance and V is the voltage. For a 2.0-μF capacitor charged to 100 V, Q = 2.0 μF × 100 V = 200 μC.

Question 8

A parallel plate capacitor with plate separation of 4.0 cm has a plate area of 6.0 × 10^-2 m². What is the capacitance of this capacitor if a dielectric material with a dielectric constant of 2.4 is placed between the plates?

Accepted Answer

The capacitance C is given by the formula C = (ε₀ * κ * A) / d, where ε₀ is the permittivity of free space (8.85 × 10^-12 F/m), κ is the dielectric constant (2.4), A is the area (6.0 × 10^-2 m²), and d is the separation (0.04 m). Calculating gives C = 32 × 10^-12 F.

Question 9

A parallel plate capacitor with plate separation of 4.0 cm has a plate area of 4.0 × 10^-2 m². What is the capacitance of this capacitor with air between these plates?

Accepted Answer

The capacitance of a parallel plate capacitor with plate area A and separation d is given by C = εA/d, where ε is the permittivity of the medium between the plates. In the case of air, ε ≈ 8.85 x 10^-12 F/m. Substituting the values given in the problem, we have C = (8.85 x 10^-12 F/m)(4.0 x 10^-2 m^2)/(4.0 x 10^-2 m) = 8.9 x 10^-12 F = 8.9 x 10^-12 F or 0.089 pF. Therefore, the best choice is B.

Question 10

FIGURE 20-11 
-Three point charges of -2.00 μC, +4.00 μC, and +6.00 μC are placed along the x-axis as shown in Figure 20-11. What is the electrical potential at point P due to these charges?

Accepted Answer

To find the electrical potential at point P, we need to calculate the contribution of each charge and then sum them up. The electrical potential at point P due to a point charge q is given by:

V = k * q / r

where k is Coulomb's constant, q is the charge, and r is the distance between the charge and the point P.

The distance between point charges and P is the same for all three charges, and it is given as:

r = 0.20 m

Using this, we can calculate the contribution of each charge as follows:

For the charge of -2.00 μC:

V1 = k * (-2.00 μC) / 0.20 m = -90 × 10^6 V

For the charge of +4.00 μC:

V2 = k * (+4.00 μC) / 0.20 m = +181 × 10^6 V

For the charge of +6.00 μC:

V3 = k * (+6.00 μC) / 0.20 m = +272 × 10^6 V

The total electrical potential at point P due to all three charges is the sum of these individual potentials:

Vtotal = V1 + V2 + V3

= (-90 + 181 + 272) × 10^6 V

= +363 × 10^6 V

= +363 × 10^3 kV

= +363 kV

So the answer is B, +307 × 10^6 V.

Question 11

FIGURE 20-11 
-The absolute value of the electric potential at a distance of 4 m from a certain point charge is 200 V. What is the absolute value of this potential at a distance of 2 m from the same charge?

Accepted Answer

The electric potential V due to a point charge is inversely proportional to the distance r from the charge, V ∝ 1/r. If the potential is 200 V at 4 m, at 2 m (which is half the distance), the potential will be twice as much, i.e., 400 V.

Question 12

FIGURE 20-10 
-Two point charges of magnitude +4.00 μC and +2.00 μC are placed at the opposite corners of a rectangle as shown in Figure 20-10. What is the potential difference between points A and B?

Accepted Answer

The answer of FIGURE 20-10 
-Two point charges of magnitude...

Question 13

A 4-μF capacitor has a potential drop of 20 V between its plates. The electric potential energy stored in this capacitor is:

Accepted Answer

The electric potential energy stored in a capacitor is given by:

U = (1/2)*C*(V^2)

where U is the electric potential energy, C is the capacitance, and V is the potential difference between the plates.

Substituting the given values, we get:

U = (1/2)*(4*10^-6)*(20)^2 = 800*10^-6 J = 800 μJ

Therefore, the answer is (D).

Question 14

FIGURE 20-13 
-The equipotential surfaces between the two point charges are shown in Figure 20-13 with the value of potential marked for each line. If the distance between points C and D is also 2.5 cm, what is the intensity of the electric field between these two points?

Accepted Answer

The answer of FIGURE 20-13 
-The equipotential surfaces between the...

Question 15

FIGURE 20-13 
-The equipotential surfaces between the two point charges are shown in Figure 20-13 with the value of potential marked for each line. If the distance between points A and B is 2.5 cm what is the intensity of the electric field between these two points?

Accepted Answer

The electric field intensity (E) is calculated using the formula E = ΔV/Δd, where ΔV is the potential difference and Δd is the distance. If the potential difference between points A and B is 0.2 V and the distance is 0.025 m (2.5 cm), then E = 0.2 V / 0.025 m = 8 V/m.

Question 16

FIGURE 20-14 
-The equipotential surfaces between two spherical conductors are shown in Figure 20-14, with the value of the potential marked for each line. What is the potential difference between points A and G?

Accepted Answer

The potential difference between two points on equipotential surfaces is the difference in their potential values. If point A is on a surface with a potential of 320 V and point G is on a surface with a potential of 0 V, the potential difference is 320 V - 0 V = 320 V.

Question 17

The magnitude of the charge on each plate of a parallel plate capacitor is 4 μC and the potential difference between the plates is 80 V. What is the capacitance of this capacitor?

Accepted Answer

Capacitance (C) is calculated using the formula C = Q/V, where Q is the charge (4 μC = 4 × 10^-6 C) and V is the potential difference (80 V). Thus, C = (4 × 10^-6 C) / (80 V) = 5 × 10^-8 F.

Question 18

A 6.00-μF parallel plate capacitor has a charge of + 40.0 μC and - 40.0 μC on each plate, respectively. The potential energy stored in this capacitor is:

Accepted Answer

The answer of A 6.00-μF parallel plate capacitor has a...

Question 19

FIGURE 20-12 
-Two point charges of magnitude +4.00 μC and -4.00 μC are placed as shown in Figure 20-12. An additional charge of magnitude +2.00 μC is placed at point A. What is the electric potential energy of this system of charges?

Accepted Answer

The electric potential energy of a system of charges is given by EPE = kq1q2/r12 + kq1q3/r13 + kq2q3/r23, where k is the Coulomb constant, q1, q2, and q3 are the charges, and r12, r13, and r23 are the distances between the charges. Plugging in the values, we get:

EPE = (9 x 10^9 Nm^2/C^2)(4 x 10^-6 C)(-4 x 10^-6 C)/0.10 m + (9 x 10^9 Nm^2/C^2)(4 x 10^-6 C)(2 x 10^-6 C)/0.20 m + (9 x 10^9 Nm^2/C^2)(-4 x 10^-6 C)(2 x 10^-6 C)/0.30 m

Simplifying, we get:

EPE = -264 x 10^-3 J

The negative sign indicates that the system is bound, meaning it takes work to separate the charges.

Quiz 20: Electric Potential and Electric Potential Energy

The potential difference between the plates of a parallel plate capacitor with the plate separation of 6 cm is 60 V. What is the electric field between the plates of this capacitor?

A parallel plate capacitor has a potential difference between the plates of 80 V. If the charge on one of the plates of the capacitor is +8.0 μC, what is the electrical energy stored by this capacitor?

FIGURE 20-12 -Two point charges of magnitude +4.0 μC and -4.0 μC are placed as shown in Figure 20-12. What is the potential difference between points A and B?

The potential difference between the plates of a parallel plate capacitor is 35 V and the electric field between the plates has a strength of 750 V/m. If the plate area is 4.0 × 10-2 m2, what is the capacitance of this capacitor?

FIGURE 20-14 -The equipotential surfaces between two spherical conductors are shown in Figure 20-14, with the value of the potential marked for each line. What is the potential difference between points G and D?

A 20-μF capacitor has a charge of 60 μC. How much energy is stored in this capacitor?

What charge appears on the plates of a 2.0-μF capacitor when it is charged to 100 V?

A parallel plate capacitor with plate separation of 4.0 cm has a plate area of 6.0 × 10-2 m2. What is the capacitance of this capacitor if a dielectric material with a dielectric constant of 2.4 is placed between the plates?

A parallel plate capacitor with plate separation of 4.0 cm has a plate area of 4.0 × 10-2 m2. What is the capacitance of this capacitor with air between these plates?

FIGURE 20-11 -Three point charges of -2.00 μC, +4.00 μC, and +6.00 μC are placed along the x-axis as shown in Figure 20-11. What is the electrical potential at point P due to these charges?

FIGURE 20-11 -The absolute value of the electric potential at a distance of 4 m from a certain point charge is 200 V. What is the absolute value of this potential at a distance of 2 m from the same charge?

FIGURE 20-10 -Two point charges of magnitude +4.00 μC and +2.00 μC are placed at the opposite corners of a rectangle as shown in Figure 20-10. What is the potential difference between points A and B?

A 4-μF capacitor has a potential drop of 20 V between its plates. The electric potential energy stored in this capacitor is:

FIGURE 20-13 -The equipotential surfaces between the two point charges are shown in Figure 20-13 with the value of potential marked for each line. If the distance between points C and D is also 2.5 cm, what is the intensity of the electric field between these two points?

FIGURE 20-13 -The equipotential surfaces between the two point charges are shown in Figure 20-13 with the value of potential marked for each line. If the distance between points A and B is 2.5 cm what is the intensity of the electric field between these two points?

FIGURE 20-14 -The equipotential surfaces between two spherical conductors are shown in Figure 20-14, with the value of the potential marked for each line. What is the potential difference between points A and G?

The magnitude of the charge on each plate of a parallel plate capacitor is 4 μC and the potential difference between the plates is 80 V. What is the capacitance of this capacitor?

A 6.00-μF parallel plate capacitor has a charge of + 40.0 μC and - 40.0 μC on each plate, respectively. The potential energy stored in this capacitor is:

FIGURE 20-12 -Two point charges of magnitude +4.00 μC and -4.00 μC are placed as shown in Figure 20-12. An additional charge of magnitude +2.00 μC is placed at point A. What is the electric potential energy of this system of charges?

The potential difference between the plates of a parallel plate capacitor is 35 V and the electric field between the plates has a strength of 750 V/m. If the plate area is 4.0 × 10^-2 m², what is the capacitance of this capacitor?

A parallel plate capacitor with plate separation of 4.0 cm has a plate area of 6.0 × 10^-2 m². What is the capacitance of this capacitor if a dielectric material with a dielectric constant of 2.4 is placed between the plates?

A parallel plate capacitor with plate separation of 4.0 cm has a plate area of 4.0 × 10^-2 m². What is the capacitance of this capacitor with air between these plates?