Answer to Question #250663 in Physics for Ramses

Question

A plate with positive charge is 0.05 m higher than a plate with negative charge, and the electric field has an intensity of E=5x10^4N/C. Determine the work performed by the electric field E when a charge of +5x10^-6C moves from the negative plate to the positive plate.

Answer 1

By definition, the work required to move charge $q = 5\times 10^{-6}C$ from A (negative plate) to B (positive plate) is given as follows:

$$W = q\phi_{AB}$$

where $\phi_{AB}$ is the potential difference between A and B. In turn, the potential difference between capacitor plates is given as follows:

$$\phi_{AB} = -Ed$$

where $E=5\times 10^4N/C$ is the electric field, and $d = 0.05m$ is the distance between plates ('-' sign since the charge moves from negative to positive). Thus, obtain:

$$W = -qEd = -5\times 10^{-6}\cdot 5\times 10^4\cdot 0.05 = -0.0125 N\cdot m$$

Answer. -0.0125 Nm.