Question

Two points, A and B, are separated by 0.0158 m. The potential at A is +81.8 V, and that at B is +40.1 V. Find the magnitude of the constant electric field between the points.

Answer 1

2639.24 V/m

Step-by-step explanation:

Electric field = potential difference between point A and B ÷ distance apart

Potential at A = 81.8 V

Potential at B = 40.1 V

Potential difference between A and B = 81.8 - 40.1 = 40.7 V

Distance apart = 0.0158 m

Electric field = 40.7 V ÷ 0.0158 m = 2639.24 V/m