A 3.6 $\mu$C charge is moved from a position where its electric potential energy is 7.2 my to a position where its potential energy is 1.8 my.What is the change in potential from the initial position to the final position?

Question

Quizplus · Accepted Answer

The electric potential difference, or voltage, between the initial and final positions can be found using the formula:

∆V = ∆PE / q

where ∆PE is the change in electric potential energy (7.2 µJ - 1.8 µJ = 5.4 µJ) and q is the charge of the object (3.6 µC).

Plugging in the values:

∆V = (5.4 µJ) / (3.6 µC) = 1500 V

The negative sign is not used for the answer because the problem only asks for the magnitude of the potential difference, not its direction.

A 36 μ\muμ C Charge Is Moved from a Position Where Its Electric

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