Answer to Question #241634 in Electric Circuits for Ali

Question

A charged pith ball of mass 0.12 kg is suspended between parallel plates. The negative plate is on the bottom, and the plates are 0.5m apart. The pith ball is located halfway between the plates. Find a) the strength and direction of the electric field, b) the sign of the charge on the pith ball, c) the potential at the location of the pith ball, and d) what would happen if the pith ball were moved to a point 0.15 m above the negative plate. Show your work and explain your reasoning.

Answer 1

a) The strength of the electric field will be equal to the balance between the electric force and the weight:

$qE-mg=0 \\ \implies E=mg/q = -mg/Q$

The electric force will be directed against the weight vector.

b) q= -Q, where Q is the charge of the plates. Since the plates had a charge Q<0, then q>0 (positive charge)

c) Since

$E=V_{ab}/d \\ \implies V_{ab}=Ed=-{mg}/{Qd}$

Since Q<0, the average potential will be V>0

d) If the charge moves to a higher position then the electric potential will diminish to counteract the force exerted by the weight.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.