The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

part a

if the charge was started from rest and had 4.72×10−4 j of kinetic energy when it reached point b, what must be the potential difference between a and b?

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge:

where

q is the charge

is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

and since the charge starts from rest, , so the formula becomes

In this problem, we have

is the work done by the external force

is the charge

is the final kinetic energy

Solving the formula for , we find

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