## Mastering Physics Solutions: Electric Fields and Equipotential Surfaces

Electric Fields and Equipotential Surfaces

Part A = 0 J
Part B = 1 J
Part C = greater than the magnitude of the electric field at point B.

Solutions Below:

The dashed lines in the diagram represent cross sections of equipotential surfaces drawn in 1 V increments.

Part A

What is the work WAB done by the electric force to move a 1 C charge from A to B?

Since points A and B are on the same equipotential, there is no change in electric potential and therefore no work needed to move the charge.

0 J

Part B

What is the work WAD done by the electric force to move a 1 C charge from A to D?

There is a 1V difference between the two equipotentials, therefore there is 1 J of work done to move the charge.

1 J

Part C

The magnitude of the electric field at point C is

• greater than the magnitude of the electric field at point B.
• less than the magnitude of the electric field at point B.
• equal to the magnitude of the electric field at point B.
• unknown because the value of the electric potential at point C is unknown.

Point C lies between the -1V and -2V equipotentials. Since B is on the 1V equipotential, the magnitude of the electric field at point C must be greater.

greater than the magnitude of the electric field at point B

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