Question 1

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on the x axis, and a 70-nc charge at y = −6 m on the y axis. What is the electric potential (relative to a zero at infinity) at the point x = 8 m on the x axis?

Accepted Answer

The electric potential $V$ at a point due to a point charge $q$ is given by $V = \frac{kq}{r}$, where $k$ is Coulomb's constant ($9 \times 10^9 \, \text{Nm}^2/\text{C}^2$) and $r$ is the distance from the charge to the point in question. For multiple charges, the total potential is the algebraic sum of the potentials due to each charge.1. For the 50-nC charge at $y = 6 \, \text{m}$, the distance to the point $x = 8 \, \text{m}$ is $\sqrt{8^2 + 6^2} = 10 \, \text{m}$. The potential due to this charge is $V_1 = \frac{(9 \times 10^9)(50 \times 10^{-9})}{10} = 45 \, \text{V}$.2. For the -80-nC charge at $x = -4 \, \text{m}$, the distance to the point $x = 8 \, \text{m}$ is $8 - (-4) = 12 \, \text{m}$. The potential due to this charge is $V_2 = \frac{(9 \times 10^9)(-80 \times 10^{-9})}{12} = -60 \, \text{V}$.3. For the 70-nC charge at $y = -6 \, \text{m}$, the distance to the point $x = 8 \, \text{m}$ is $\sqrt{8^2 + 6^2} = 10 \, \text{m}$. The potential due to this charge is $V_3 = \frac{(9 \times 10^9)(70 \times 10^{-9})}{10} = 63 \, \text{V}$.The total potential at $x = 8 \, \text{m}$ is the sum of these potentials: $V_{\text{total}} = V_1 + V_2 + V_3 = 45 - 60 + 63 = 48 \, \text{V}$.

Question 2

Point charges of equal magnitudes (25 nC) and opposite signs are placed on (diagonally) opposite corners of a 60-cm × 80-cm rectangle. If point A is the corner of this rectangle nearest the positive charge and point B is located at the intersection of the diagonals of the rectangle, determine the potential difference, V_B − V_A.

Accepted Answer

D

Question 3

Points A [at (3, 6) m] and B [at (8, −3) m] are in a region where the electric field is uniform and given by N/C. What is the electric potential difference V_A − V_B?

Accepted Answer

A

Question 4

Identical 2.0-μC charges are located on the vertices of a square with sides that are 2.0 m in length. Determine the electric potential (relative to zero at infinity) at the center of the square.

Accepted Answer

The answer of Identical 2.0-μC charges are located on the...

Question 5

A particle (mass = 6.7 × 10−²⁷ kg, charge = 3.2 × 10−¹⁹ C) moves along the positive x axis with a speed of 4.8 × 10⁵ m/s. It enters a region of uniform electric field parallel to its motion and comes to rest after moving 2.0 m into the field. What is the magnitude of the electric field?

Accepted Answer

The initial kinetic energy of the particle is converted into potential energy as it moves into the electric field. The work done by the electric field on the particle is equal to the change in potential energy. Thus, we can use the conservation of energy to find the electric field strength:

(1/2)mv^2 = qEd

where m is the mass of the particle, v is its initial velocity, q is its charge, E is the electric field strength, and d is the distance it travels in the electric field before coming to rest.

Solving for E, we get:

E = (1/2)(mv^2)/qd

Substituting the given values, we get:

E = (1/2)(6.7 × 10^-31 kg)(4.8 × 10^5 m/s)^2/(3.2 × 10^-19 C)(2.0 m)

E ≈ 1.2 kN/C

Therefore, the answer is choice C, 1.2 kN/C.

Question 6

If a = 30 cm, b = 20 cm, q = +2.0 nC, and Q = −3.0 nC in the figure, what is the potential difference V_A − V_B?

Accepted Answer

The answer of If a = 30 cm, b =...

Question 7

A +4.0-μC charge is placed on the x axis at x = +3.0 m, and a −2.0-μC charge is located on the y axis at y = −1.0 m. Point A is on the y axis at y = +4.0 m. Determine the electric potential at point A (relative to zero at the origin).

Accepted Answer

The electric potential at a point due to a point charge is given by V = k * q / r, where k is Coulomb's constant (8.99 x 10^9 N m²/C²), q is the charge, and r is the distance from the charge to the point. Calculate the potential at point A due to each charge separately and sum them. The distance from the +4.0 μC charge to point A is √((3.0 m)^2 + (4.0 m)^2) = 5.0 m, and from the -2.0 μC charge to point A is 5.0 m. The total potential is V = (8.99 x 10^9)(4.0 x 10^-6)/5.0 + (8.99 x 10^9)(-2.0 x 10^-6)/5.0 = 9.6 kV.

Question 8

A particle (charge = 50 μC) moves in a region where the only force on it is an electric force. As the particle moves 25 cm from point A to point B, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference, V_B − V_A.

Accepted Answer

The answer of A particle (charge = 50 μC) moves...

Question 9

An alpha particle (m = 6.7 × 10−²⁷ kg, q = +3.2 × 10−¹⁹ C) has a speed of 20 km/s at point A and moves to point B where it momentarily stops. Only electric forces act on the particle during this motion. Determine the electric potential difference V_A − V_B.

Accepted Answer

The answer of An alpha particle (m = 6.7 ×...

Question 10

A charged particle (q = −8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is released from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference V_B − V_A?

Accepted Answer

Using conservation of energy, we have:
qVS1U1B0 - qVS1U1B1 - K = 0
Where K is the kinetic energy of the charged particle at point B.
Solving for VS1U1B1, we have:
qVS1U1B1 = qVS1U1B0 - K
VS1U1B1 = VS1U1B0 - K/q
Note that q is negative, so we have:
VS1U1B1 - VS1U1B0 = - K/|q|
Substituting K = 4.8 J and q = -8.0x10^-6 C, we get:
VS1U1B1 - VS1U1B0 = - 4.8 J/|(-8.0x10^-6 C)| = +0.60 kV
Therefore, the answer is B: +0.60 kV.

Question 11

Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an electric field of  N/C at point P. Determine the work required of an external agent to move a 3.0-μC charge along the x axis from infinity to point P without any net change in the kinetic energy of the particle.

Accepted Answer

The answer of Several charges in the neighborhood of point...

Question 12

A proton (m = 1.7 × 10−²⁷ kg, q = +1.6 × 10−¹⁹ C) starts from rest at point A and has a speed of 40 km/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference V_B − V_A.

Accepted Answer

The electric potential difference is calculated using the kinetic energy change: ΔKE = q(VB - VA). The change in kinetic energy is (1/2)mv^2. Solving for VB - VA gives -8.5 V.

Question 13

A particle (m = 2.0 μg, q = −5.0 μC) has a speed of 30 m/s at point A and moves (with only electric forces acting on it) to point B where its speed is 80 m/s. Determine the electric potential difference V_A − V_B.

Accepted Answer

The change in kinetic energy of the particle is equal to the work done by the electric field, which is also equal to the charge times the electric potential difference (ΔV = ΔKE/q). The initial kinetic energy (KE1) at point A is (1/2)mv^2 = (1/2)(2.0 x 10^-6 kg)(30 m/s)^2 = 9 x 10^-4 J. The final kinetic energy (KE2) at point B is (1/2)mv^2 = (1/2)(2.0 x 10^-6 kg)(80 m/s)^2 = 6.4 x 10^-3 J. The change in kinetic energy (ΔKE) is KE2 - KE1 = 6.4 x 10^-3 J - 9 x 10^-4 J = 5.5 x 10^-3 J. Since q = -5.0 μC = -5.0 x 10^-6 C, the electric potential difference is ΔV = ΔKE/q = (5.5 x 10^-3 J) / (-5.0 x 10^-6 C) = -1.1 x 10^3 V = -1.1 kV.

Question 14

A particle (charge = +2.0 mC) moving in a region where only electric forces act on it has a kinetic energy of 5.0 J at point A. The particle subsequently passes through point B which has an electric potential of +1.5 kV relative to point A. Determine the kinetic energy of the particle as it moves through point B.

Accepted Answer

The answer of A particle (charge = +2.0 mC) moving...

Question 15

What is the speed of a proton that has been accelerated from rest through a potential difference of 4.0 kV?

Accepted Answer

The kinetic energy gained by the proton is equal to the electrical energy provided by the potential difference, which can be calculated using the formula $KE = qV$, where $q$ is the charge of the proton (approximately $1.6 \times 10^{-19}$ C) and $V$ is the potential difference (4.0 kV or $4.0 \times 10^3$ V). The kinetic energy is also given by $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the proton ($1.67 \times 10^{-27}$ kg) and $v$ is the velocity. Equating the two expressions for kinetic energy and solving for $v$ gives the speed of the proton. The correct answer is the one that matches this calculation, which is $8.8 \times 10^{6}$ m/s.

Question 16

A proton (mass = 1.67 × 10−²⁷ kg, charge = 1.60 × 10−¹⁹ C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 50 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference V_B − V_A.

Accepted Answer

The answer of A proton (mass = 1.67 × 10−²⁷...

Question 17

An electron (m = 9.1 × 10−³¹ kg, q = −1.6 × 10−¹⁹ C) starts from rest at point A and has a speed of 5.0 × 10⁶ m/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference V_A − V_B.

Accepted Answer

The electric potential difference is given by:$$V_A - V_B = \frac{1}{2}mv_B^2$$$$V_A - V_B = \frac{1}{2}(9.1 	imes 10^{-31} 	ext{ kg})(5.0 	imes 10^6 	ext{ m/s})^2$$$$V_A - V_B = -71 	ext{ V}$$Therefore, the correct answer is A.

Question 18

Point charges q and Q are positioned as shown. If q = +2.0 nC, Q = −2.0 nC, a = 3.0 m, and b = 4.0 m, what is the electric potential difference, V_A − V_B?

Accepted Answer

First, we need to find the electric potential at point A and point B due to the point charge Q, using the formula:

V = kQ/r

where k is the Coulomb constant (9x10^9 Nm^2/C^2), Q is the charge, and r is the distance from the charge to the point where the potential is being calculated.

At point A: 
V1 = kQ/a = (9x10^9)(-2x10^-9)/3 = -6 V

At point B: 
V0 = kQ/b = (9x10^9)(-2x10^-9)/4 = -4.5 V

Next, we need to find the electric potential at point A and point B due to the point charge q, using the same formula.

At point A: 
VS1U1B1AS1U1B0 = kq/SA 
where SA is the distance from point A to the point where VS1U1B1AS1U1B0 is being calculated, which is 1 m (since it is equidistant to point S1 and point U1).

VS1U1B1AS1U1B0 = kq/SA = (9x10^9)(2x10^-9)/1 = 18 V

At point B: 
VS1U1B1BS1U1B0 = kq/SB = (9x10^9)(2x10^-9)/5 = 3.6 V

Finally, we can find the electric potential difference between point A and point B as:

VS1U1B1AS1U1B0 - VS1U1B1BS1U1B0 = 18 V - 3.6 V = 14.4 V

However, the question is asking for the potential difference between points S1, U1, and B0, not between points A and B. We can use the fact that the potential difference between two points in a conservative electric field is independent of the path taken between those points. Therefore, we can calculate the potential difference between points S1, U1, and B0 as the sum of the potential differences between adjacent points:

VS1U1B1AS1U1B0 - VS1U1B1BS1U1B0 = (VS1U1B1AS1U1B1 - VS1U1B0AS1U1B0) + (VS1U1B0AS1U1B0 - VS1U1B1BS1U1B0)

The potential difference between S1 and U1 due to the charge q is:

VS1U1B0AS1U1B0 = kq/r 
where r is the distance from the charge to the midpoint between points S1 and U1, which is 1.5 m.

VS1U1B0AS1U1B0 = (9x10^9)(2x10^-9)/1.5 = 12 V

The potential difference between U1 and B0 due to the charge Q is:

VU1B0 = kQ/r 
where r is the distance from the charge to point B0, which is 4 m.

VU1B0 = (9x10^9)(-2x10^-9)/4 = -4.5 V

Therefore, the potential difference between points S1, U1, and B0 is:

14.4 V = (18 V - 12 V) + (12 V - (-4.5 V)) = 6 V

Therefore, the answer is B) 6.0 V.

Question 19

A proton (mass = 1.67 × 10−²⁷ kg, charge = 1.60 × 10−¹⁹ C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 60 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference V_B − V_A.

Accepted Answer

The answer of A proton (mass = 1.67 × 10−²⁷...

Question 20

Points A [at (2, 3) m] and B [at (5, 7) m] are in a region where the electric field is uniform and given by N/C. What is the potential difference V_A − V_B?

Accepted Answer

The answer of Points A [at (2, 3) m] and...

Quiz 25: Electric Potential

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on the x axis, and a 70-nc charge at y = −6 m on the y axis. What is the electric potential (relative to a zero at infinity) at the point x = 8 m on the x axis?

Points A [at (3, 6) m] and B [at (8, −3) m] are in a region where the electric field is uniform and given by N/C. What is the electric potential difference V_A − V_B?

Identical 2.0-μC charges are located on the vertices of a square with sides that are 2.0 m in length. Determine the electric potential (relative to zero at infinity) at the center of the square.

If a = 30 cm, b = 20 cm, q = +2.0 nC, and Q = −3.0 nC in the figure, what is the potential difference V_A − V_B?

A +4.0-μC charge is placed on the x axis at x = +3.0 m, and a −2.0-μC charge is located on the y axis at y = −1.0 m. Point A is on the y axis at y = +4.0 m. Determine the electric potential at point A (relative to zero at the origin) .

A particle (charge = 50 μC) moves in a region where the only force on it is an electric force. As the particle moves 25 cm from point A to point B, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference, V_B − V_A.

An alpha particle (m = 6.7 × 10−²⁷ kg, q = +3.2 × 10−¹⁹ C) has a speed of 20 km/s at point A and moves to point B where it momentarily stops. Only electric forces act on the particle during this motion. Determine the electric potential difference V_A − V_B.

A charged particle (q = −8.0 mC) , which moves in a region where the only force acting on the particle is an electric force, is released from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference V_B − V_A?

A proton (m = 1.7 × 10−²⁷ kg, q = +1.6 × 10−¹⁹ C) starts from rest at point A and has a speed of 40 km/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference V_B − V_A.

A particle (m = 2.0 μg, q = −5.0 μC) has a speed of 30 m/s at point A and moves (with only electric forces acting on it) to point B where its speed is 80 m/s. Determine the electric potential difference V_A − V_B.

What is the speed of a proton that has been accelerated from rest through a potential difference of 4.0 kV?

A proton (mass = 1.67 × 10−²⁷ kg, charge = 1.60 × 10−¹⁹ C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 50 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference V_B − V_A.

An electron (m = 9.1 × 10−³¹ kg, q = −1.6 × 10−¹⁹ C) starts from rest at point A and has a speed of 5.0 × 10⁶ m/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference V_A − V_B.

Point charges q and Q are positioned as shown. If q = +2.0 nC, Q = −2.0 nC, a = 3.0 m, and b = 4.0 m, what is the electric potential difference, V_A − V_B?

A proton (mass = 1.67 × 10−²⁷ kg, charge = 1.60 × 10−¹⁹ C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 60 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference V_B − V_A.

Points A [at (2, 3) m] and B [at (5, 7) m] are in a region where the electric field is uniform and given by N/C. What is the potential difference V_A − V_B?

Quiz 25: Electric Potential

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on the x axis, and a 70-nc charge at y = −6 m on the y axis. What is the electric potential (relative to a zero at infinity) at the point x = 8 m on the x axis?

Points A [at (3, 6) m] and B [at (8, −3) m] are in a region where the electric field is uniform and given by N/C. What is the electric potential difference VA − VB?

Identical 2.0-μC charges are located on the vertices of a square with sides that are 2.0 m in length. Determine the electric potential (relative to zero at infinity) at the center of the square.

A particle (mass = 6.7 × 10−27 kg, charge = 3.2 × 10−19 C) moves along the positive x axis with a speed of 4.8 × 105 m/s. It enters a region of uniform electric field parallel to its motion and comes to rest after moving 2.0 m into the field. What is the magnitude of the electric field?

If a = 30 cm, b = 20 cm, q = +2.0 nC, and Q = −3.0 nC in the figure, what is the potential difference VA − VB?

A +4.0-μC charge is placed on the x axis at x = +3.0 m, and a −2.0-μC charge is located on the y axis at y = −1.0 m. Point A is on the y axis at y = +4.0 m. Determine the electric potential at point A (relative to zero at the origin) .

A particle (charge = 50 μC) moves in a region where the only force on it is an electric force. As the particle moves 25 cm from point A to point B, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference, VB − VA.

An alpha particle (m = 6.7 × 10−27 kg, q = +3.2 × 10−19 C) has a speed of 20 km/s at point A and moves to point B where it momentarily stops. Only electric forces act on the particle during this motion. Determine the electric potential difference VA − VB.

A charged particle (q = −8.0 mC) , which moves in a region where the only force acting on the particle is an electric force, is released from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference VB − VA?

A proton (m = 1.7 × 10−27 kg, q = +1.6 × 10−19 C) starts from rest at point A and has a speed of 40 km/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference VB − VA.

A particle (m = 2.0 μg, q = −5.0 μC) has a speed of 30 m/s at point A and moves (with only electric forces acting on it) to point B where its speed is 80 m/s. Determine the electric potential difference VA − VB.

What is the speed of a proton that has been accelerated from rest through a potential difference of 4.0 kV?

A proton (mass = 1.67 × 10−27 kg, charge = 1.60 × 10−19 C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 50 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference VB − VA.

An electron (m = 9.1 × 10−31 kg, q = −1.6 × 10−19 C) starts from rest at point A and has a speed of 5.0 × 106 m/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference VA − VB.

Point charges q and Q are positioned as shown. If q = +2.0 nC, Q = −2.0 nC, a = 3.0 m, and b = 4.0 m, what is the electric potential difference, VA − VB?

A proton (mass = 1.67 × 10−27 kg, charge = 1.60 × 10−19 C) moves from point A to point B under the influence of an electrostatic force only. At point A the proton moves with a speed of 60 km/s. At point B the speed of the proton is 80 km/s. Determine the potential difference VB − VA.

Points A [at (2, 3) m] and B [at (5, 7) m] are in a region where the electric field is uniform and given by N/C. What is the potential difference VA − VB?

Points A [at (3, 6) m] and B [at (8, −3) m] are in a region where the electric field is uniform and given by N/C. What is the electric potential difference V_A − V_B?

If a = 30 cm, b = 20 cm, q = +2.0 nC, and Q = −3.0 nC in the figure, what is the potential difference V_A − V_B?

A particle (charge = 50 μC) moves in a region where the only force on it is an electric force. As the particle moves 25 cm from point A to point B, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference, V_B − V_A.

An alpha particle (m = 6.7 × 10−²⁷ kg, q = +3.2 × 10−¹⁹ C) has a speed of 20 km/s at point A and moves to point B where it momentarily stops. Only electric forces act on the particle during this motion. Determine the electric potential difference V_A − V_B.

A charged particle (q = −8.0 mC) , which moves in a region where the only force acting on the particle is an electric force, is released from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference V_B − V_A?

A proton (m = 1.7 × 10−²⁷ kg, q = +1.6 × 10−¹⁹ C) starts from rest at point A and has a speed of 40 km/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference V_B − V_A.

A particle (m = 2.0 μg, q = −5.0 μC) has a speed of 30 m/s at point A and moves (with only electric forces acting on it) to point B where its speed is 80 m/s. Determine the electric potential difference V_A − V_B.

An electron (m = 9.1 × 10−³¹ kg, q = −1.6 × 10−¹⁹ C) starts from rest at point A and has a speed of 5.0 × 10⁶ m/s at point B. Only electric forces act on it during this motion. Determine the electric potential difference V_A − V_B.

Point charges q and Q are positioned as shown. If q = +2.0 nC, Q = −2.0 nC, a = 3.0 m, and b = 4.0 m, what is the electric potential difference, V_A − V_B?

Points A [at (2, 3) m] and B [at (5, 7) m] are in a region where the electric field is uniform and given by N/C. What is the potential difference V_A − V_B?