Two large equipotent surfaces are flat and parallel and separated by a relatively small distance compared to their size.One surface is at 25.0 V and the other is at 33.0 V.The surfaces are 1.25 cm apart.What is the magnitude of the electric field midway between the surfaces?

Question

Quizplus · Accepted Answer

The electric field between two parallel equipotential surfaces is given by:E = (V2 - V1) / dwhere:E is the electric field in V/mV2 is the potential of the higher potential surface in VV1 is the potential of the lower potential surface in Vd is the distance between the surfaces in mIn this problem, we have:V2 = 33.0 VV1 = 25.0 Vd = 1.25 cm = 0.0125 mSubstituting these values into the equation, we get:E = (33.0 V - 25.0 V) / 0.0125 m = 6.40 * 10^2 V/mTherefore, the magnitude of the electric field midway between the surfaces is 6.40 * 10^2 V/m.

Two Large Equipotent Surfaces Are Flat and Parallel and Separated ×\times×

Two Large Equipotent Surfaces Are Flat and Parallel and Separated $\times$