Question 1

An electron is carried from the positive terminal to the negative terminal of a 9 V battery. How much work is required in carrying this electron?

Accepted Answer

The work done is calculated using the formula W = qV, where q is the charge of the electron (1.6 × 10^-19 C) and V is the voltage (9 V). Thus, W = 1.6 × 10^-19 C × 9 V = 14.4 × 10^-19 J.

Question 2

A + 5.0-μC charge is moved from a negative to a positive plate of a parallel plate capacitor. In moving this charge 0.30 mJ of energy is used. What is the potential difference between the plates of this capacitor?

Accepted Answer

The potential difference (V) can be calculated using the formula V = W/Q, where W is the work done (0.30 mJ = 0.30 x 10^-3 J) and Q is the charge (5.0 μC = 5.0 x 10^-6 C). V = (0.30 x 10^-3 J) / (5.0 x 10^-6 C) = 60 V.

Question 3

FIGURE 20-3 
-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 20-3.
(a) What is the potential at point A due to these charges?
(b) What is the potential at point B due to these charges?
(c) What is the potential difference between points A and B?

Accepted Answer

The answer of FIGURE 20-3 
-Two point charges of magnitude...

Question 4

At a certain point in space there is a potential of 400 V. What is the potential energy of a +2-μC charge at that point in space?

Accepted Answer

The potential energy of a charge is given by the formula U = qV, where q is the charge and V is the potential. Plugging in the given values, we get:
U = (2 μC)(400 V)
U = 800 μJ = 8 × 10^-4 J = 8 × 10^-14 J (in scientific notation with proper units)
Therefore, the answer is (E).

Question 5

FIGURE 20-5 
-Four equal point charges of magnitude 6.00 μC and of varying signs are placed at the corners of a square 2.00 m on each side, as shown in Figure 20-5. What is the electric potential at the center of this square due to these charges?

Accepted Answer

Since the square is symmetrical and the charges are equal in magnitude, the electric fields due to opposite charges cancel each other out at the center of the square. Therefore, the electric potential at the center of the square is zero.

Question 6

The potential energy at x = 8 m is -2000 V and at x = 2 m is +400 V. What is the magnitude and direction of the electric field?

Accepted Answer

We can use the formula for electric potential difference to find the electric field.
ΔV = -Ed
where ΔV is the difference in potential energy, E is the electric field, and d is the displacement.
ΔV = Vf - Vi
where Vf is the final potential energy and Vi is the initial potential energy.
ΔV = (-2000 V) - (+400 V) = -2400 V
d = xf - xi = 8 m - 2 m = 6 m

Substituting the values in the first equation:
-2400 V = -Ed(6m)
E = 400 V/m directed parallel to the +x-axis

Question 7

At a certain point in space the electric potential is 20 V. A 4.0-μC charge is brought from infinity to that point. What is the electric potential energy of this charge at that point?

Accepted Answer

The electric potential energy of a point charge in an electric field is given by:

U = qV

where q is the charge and V is the electric potential at that point.

In this case, the charge q is 4.0-μC and the electric potential at that point is 20 V.

Substituting the values in the equation:

U = (4.0 × 10^-6 C) × (20 V) = 80 × 10^-6 J = 80 μJ

Therefore, the electric potential energy of the charge at that point is 80 μJ, which is option B.

Question 8

FIGURE 20-4 
-Four equal point charges of magnitude 6.00 μC are placed at the corners of a square 2.00 m on each side, as shown in Figure 20-4. What is the electric potential of these charges at the center of this square?

Accepted Answer

The answer of FIGURE 20-4 
-Four equal point charges of...

Question 9

A 2000 V/m electric field is directed along the +x-axis. If the potential at x = 10 m is 800 V, what is the potential at x = 6 m?

Accepted Answer

The potential difference is given by V = Ed, where E is the electric field and d is the distance. The potential decreases in the direction of the electric field. So, V(6 m) = V(10 m) + E*(10 m - 6 m) = 800 V + 2000 V/m * 4 m = 8800 V.

Question 10

The electric potential at the origin of an xy-coordinate system is 40 V. A -8.0-μC charge is brought from x = +∞ to that point. What is the electric potential energy of this charge at the origin?

Accepted Answer

The electric potential energy (U) is given by U = qV, where q is the charge and V is the potential. Here, U = (-8.0 × 10^-6 C)(40 V) = -3.2 × 10^-4 J.

Question 11

A charged particle with charge + 5.0 μC is initially at rest. It is accelerated through a potential difference of 500 V. What is the kinetic energy of this charged particle?

Accepted Answer

The kinetic energy gained by the charged particle is equal to the work done on it by the electric field. The formula for work done by an electric field is:
W = qV
where W is work done, q is charge and V is potential difference.
Substituting the values given in the question, we get:
W = (5.0 μC)(500 V) = 2.5 × 10^−3 J
Since the particle was initially at rest, all the work done on it has been converted into kinetic energy. Therefore, the kinetic energy of the charged particle is:
KE = 2.5 × 10^−3 J
which is equivalent to answer choice B.

Question 12

Two point charges of magnitude 4.0 μC and -4.0 μC are situated along the x-axis at x₁ = 2.0 m and x₂ = -2.0 m, respectively. What is the electric potential energy of this system of charges?

Accepted Answer

The electric potential energy (U) of a system of two point charges is given by the formula U = k * (q1 * q2) / r, where k is Coulomb's constant (8.99 x 10^9 N m²/C²), q1 and q2 are the charges, and r is the distance between them. Here, q1 = 4.0 μC, q2 = -4.0 μC, and r = 4.0 m. Substituting these values, U = (8.99 x 10^9) * (4.0 x 10^-6) * (-4.0 x 10^-6) / 4.0 = -36 mJ.

Question 13

An 800 V/m electric field is directed along the +x-axis. If the potential at x = 0 m is 2000 V, what is the potential at x = 2 m?

Accepted Answer

The potential difference between two points is given by the electric field times the distance between the points. In this case, the electric field is 800 V/m and the distance between the points is 2 m, so the potential difference is 800 V/m * 2 m = 1600 V. The potential at x = 0 m is 2000 V, so the potential at x = 2 m is 2000 V + 1600 V = 3600 V.

Question 14

An electron is initially at rest. It is accelerated through a potential difference of 400 V. What is the kinetic energy of this electron?

Accepted Answer

The kinetic energy gained by an electron accelerated through a potential difference V is given by the equation KE = eV, where e is the charge of the electron (1.6 × 10^-19 C). Thus, KE = 1.6 × 10^-19 C × 400 V = 6.4 × 10^-17 J.

Question 15

Two equal point charges of magnitude 4.0 μC are situated along the x-axis at x₁ = -2.0 m and x₂ = -2.0 m. What is the electric potential energy of this system of charges?

Accepted Answer

The answer of Two equal point charges of magnitude 4.0...

Question 16

An electron, initially at rest is accelerated through a potential difference of 550 V. What is the speed of the electron due to this potential difference?

Accepted Answer

We can use the equation for the kinetic energy gained by a single electron:

KE = qV = (1.602 x 10^-19 C)(550 V) = 8.81 x 10^-17 J

Since the electron starts at rest, all of this energy goes into its kinetic energy:

KE = 0.5mv^2

Solving for v gives:

v = sqrt((2KE)/m) = sqrt((2(8.81 x 10^-17 J))/(9.11 x 10^-31 kg)) = 1.39 x 10^6 m/s

Therefore, the answer is (E) 13.9 x 10^6 m/s.

Question 17

Three charges are placed as follows along the x- and y-axes of an xy-coordinate system: q₁ = 2.00 μC at x₁ = 0 m, q₂ = 4.00 μC at x₂ = 3.00 m, and q₃ = 6.00 μC at y = 4.00 m. What is the electric potential energy of this system of charges?

Accepted Answer

The electric potential energy of the system of charges is given by the formula:

U = k(q1*q2/r12 + q1*q3/r13 + q2*q3/r23)

where k is the Coulomb constant (9.0 x 10^9 N*m^2/C^2), q1 to q3 are the charges, and r12 to r23 are the distances between the charges.

Substituting the given values:

U = (9.0 x 10^9 N*m^2/C^2)(2.00 μC)(4.00 μC)/(3.00 m) + (9.0 x 10^9 N*m^2/C^2)(2.00 μC)(6.00 μC)/(4.00 m) + (9.0 x 10^9 N*m^2/C^2)(4.00 μC)(6.00 μC)/(5.00 m)

U = 94.2 mJ

Therefore, the best choice is B, 94.2 mJ.

Question 18

Three charges are placed as follows along the x- and y-axis of an xy-coordinate system: q₁ = 2.00 μC at x₁ = 0 m, q₂ = -4.00 μC at x₂ = 3.00 m, and q₃ = 6.00 μC at y = 4.00 m. 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 the equation:$$U = k\frac{q_1q_2}{r_{12}} + k\frac{q_1q_3}{r_{13}} + k\frac{q_2q_3}{r_{23}}$$where k is Coulomb's constant, q₁, q₂, and q₃ are the charges of the three particles, and r₁₂, r₁₃, and r₂₃ are the distances between the particles.In this problem, we have:$$q_1 = 2.00 \mu C = 2.00 imes 10^{-6} C$$$$q_2 = -4.00 \mu C = -4.00 imes 10^{-6} C$$$$q_3 = 6.00 \mu C = 6.00 imes 10^{-6} C$$$$r_{12} = 3.00 m$$$$r_{13} = \sqrt{4.00^2 + 3.00^2} m = 5.00 m$$$$r_{23} = 4.00 m$$Substituting these values into the equation for U, we get:$$U = (9.00 imes 10^9 N m^2/C^2)\frac{(2.00 imes 10^{-6} C)(-4.00 imes 10^{-6} C)}{3.00 m} + (9.00 imes 10^9 N m^2/C^2)\frac{(2.00 imes 10^{-6} C)(6.00 imes 10^{-6} C)}{5.00 m} + (9.00 imes 10^9 N m^2/C^2)\frac{(-4.00 imes 10^{-6} C)(6.00 imes 10^{-6} C)}{4.00 m}$$$$U = -40.2 mJ$$Therefore, the electric potential energy of this system of charges is -40.2 mJ.

Question 19

A proton falls through a potential drop of 400 V. How much is the change of potential energy of this proton in falling through this potential drop?

Accepted Answer

The answer of A proton falls through a potential drop...

Question 20

The potential difference between the plates of a parallel plate capacitor is 4.00 V. If the plate separation for this capacitor is 6.00 cm, what is the intensity of the electric field between the plates of this capacitor?

Accepted Answer

The answer of The potential difference between the plates of...

Quiz 20: Electric Potential and Electric Potential Energy