Net work done, and force exerted in moving a test charge from one point to the other if points are at the same potential?

Question

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Answer 1

Question

Chapter 17, Problem 1Q

To determine

Net work done, and force exerted in moving a test charge from one point to the other if points are at the same potential?

Expert Solution & Answer

Answer to Problem 1Q

Solution:

Net work done = 0

Force exerted may or may not be zero.

Explanation of Solution

We have potential difference between two points, $V_{b a} = V_{b} - V_{a} = - \frac{W_{b a}}{q}$ , where $W_{b a}$ is the net work done and q is the charge. Here $V_{b} - V_{a} = 0$ , so we will get net work done $W_{b a} = 0$ . So net work done in moving a charge from one point to other if the points are at the same potential is zero.

If the charge is taken to a higher potential first and then to the earlier potential, in first step there will be a negative work and in second step there is a positive work. But the net work done is zero.

Force exerted on a charge when it is taken along an equipotential line is zero. But when there is negative work and positive work there will be a force exerted.

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Answer 2

Question

Chapter 17, Problem 1Q

To determine

Net work done, and force exerted in moving a test charge from one point to the other if points are at the same potential?

Expert Solution & Answer

Answer to Problem 1Q

Solution:

Net work done = 0

Force exerted may or may not be zero.

Explanation of Solution

We have potential difference between two points, $V_{b a} = V_{b} - V_{a} = - \frac{W_{b a}}{q}$ , where $W_{b a}$ is the net work done and q is the charge. Here $V_{b} - V_{a} = 0$ , so we will get net work done $W_{b a} = 0$ . So net work done in moving a charge from one point to other if the points are at the same potential is zero.

If the charge is taken to a higher potential first and then to the earlier potential, in first step there will be a negative work and in second step there is a positive work. But the net work done is zero.

Force exerted on a charge when it is taken along an equipotential line is zero. But when there is negative work and positive work there will be a force exerted.

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Answer 3

Question

Chapter 17, Problem 1Q

To determine

Net work done, and force exerted in moving a test charge from one point to the other if points are at the same potential?

Expert Solution & Answer

Answer to Problem 1Q

Solution:

Net work done = 0

Force exerted may or may not be zero.

Explanation of Solution

We have potential difference between two points, $V_{b a} = V_{b} - V_{a} = - \frac{W_{b a}}{q}$ , where $W_{b a}$ is the net work done and q is the charge. Here $V_{b} - V_{a} = 0$ , so we will get net work done $W_{b a} = 0$ . So net work done in moving a charge from one point to other if the points are at the same potential is zero.

If the charge is taken to a higher potential first and then to the earlier potential, in first step there will be a negative work and in second step there is a positive work. But the net work done is zero.

Force exerted on a charge when it is taken along an equipotential line is zero. But when there is negative work and positive work there will be a force exerted.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 17, Problem 1Q

To determine

Net work done, and force exerted in moving a test charge from one point to the other if points are at the same potential?

Expert Solution & Answer

Answer to Problem 1Q

Solution:

Net work done = 0

Force exerted may or may not be zero.

Explanation of Solution

We have potential difference between two points, $V_{b a} = V_{b} - V_{a} = - \frac{W_{b a}}{q}$ , where $W_{b a}$ is the net work done and q is the charge. Here $V_{b} - V_{a} = 0$ , so we will get net work done $W_{b a} = 0$ . So net work done in moving a charge from one point to other if the points are at the same potential is zero.

If the charge is taken to a higher potential first and then to the earlier potential, in first step there will be a negative work and in second step there is a positive work. But the net work done is zero.

Force exerted on a charge when it is taken along an equipotential line is zero. But when there is negative work and positive work there will be a force exerted.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 17, Problem 1Q

To determine

Net work done, and force exerted in moving a test charge from one point to the other if points are at the same potential?

Expert Solution & Answer

Answer to Problem 1Q

Solution:

Net work done = 0

Force exerted may or may not be zero.

Explanation of Solution

We have potential difference between two points, $V_{b a} = V_{b} - V_{a} = - \frac{W_{b a}}{q}$ , where $W_{b a}$ is the net work done and q is the charge. Here $V_{b} - V_{a} = 0$ , so we will get net work done $W_{b a} = 0$ . So net work done in moving a charge from one point to other if the points are at the same potential is zero.

If the charge is taken to a higher potential first and then to the earlier potential, in first step there will be a negative work and in second step there is a positive work. But the net work done is zero.

Force exerted on a charge when it is taken along an equipotential line is zero. But when there is negative work and positive work there will be a force exerted.

Answer 6

Chapter 17, Problem 1Q

To determine

Net work done, and force exerted in moving a test charge from one point to the other if points are at the same potential?

Expert Solution & Answer

Answer to Problem 1Q

Solution:

Net work done = 0

Force exerted may or may not be zero.

Explanation of Solution

We have potential difference between two points, $V_{b a} = V_{b} - V_{a} = - \frac{W_{b a}}{q}$ , where $W_{b a}$ is the net work done and q is the charge. Here $V_{b} - V_{a} = 0$ , so we will get net work done $W_{b a} = 0$ . So net work done in moving a charge from one point to other if the points are at the same potential is zero.

If the charge is taken to a higher potential first and then to the earlier potential, in first step there will be a negative work and in second step there is a positive work. But the net work done is zero.

Force exerted on a charge when it is taken along an equipotential line is zero. But when there is negative work and positive work there will be a force exerted.

Answer 7

Chapter 17, Problem 1Q

To determine

Net work done, and force exerted in moving a test charge from one point to the other if points are at the same potential?

Answer 8

Expert Solution & Answer

Answer to Problem 1Q

Solution:

Net work done = 0

Force exerted may or may not be zero.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

We have potential difference between two points, $V_{b a} = V_{b} - V_{a} = - \frac{W_{b a}}{q}$ , where $W_{b a}$ is the net work done and q is the charge. Here $V_{b} - V_{a} = 0$ , so we will get net work done $W_{b a} = 0$ . So net work done in moving a charge from one point to other if the points are at the same potential is zero.

If the charge is taken to a higher potential first and then to the earlier potential, in first step there will be a negative work and in second step there is a positive work. But the net work done is zero.

Force exerted on a charge when it is taken along an equipotential line is zero. But when there is negative work and positive work there will be a force exerted.