Question

Consider two parallel plates (as in a parallel plate capacitor), with plates which are very much wider than their separation, with an electric potential across the plates. Sketch the equipotential lines and field lines between the plates.

If, the distance between the plates is 0.10 m, the potential of one plate is 0 V and the potential of the second plate is 10 V, what is the potential at a point 0.02 m from the second plate?

Answer 1

Equipotential surfaces are presented by dotted lines that are always drawn at right angles to the field lines (the solid vectors). Field lines always point from regions of high potential to regions of low potential. In diagram below, the top plate would be at +10 V and is the "high potential plate" while the bottom plate would be at 0 V and is the "low potential plate."

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ΔV=E⋅d
where ΔV = Change in potential.
d = Distance between the plate
E = − ΔV/d
E = -(10) 0.10
E = 100 V/m

ΔV = − Ed cos θ
ΔV = − 100 * 0.02 cos (0^o)
ΔV = − 2v

Potential at 0.02 m from the second plate = 10v -2v = 8v