Question

A small 12.6 g plastic ball is tied to a very light 28.6 cm string that is attached to vertical wall of room (see figure). A uniform horizontal electric field exists in this room. When the ball has been given an excess charge of - 1.33 μC, you observe that it remains suspended, with the string making an angle of 17.4° with the wall. Find the magnitude of the electric field in the room.

Answer 1

forces acting on the charge

1) electric force

2) tension

3) weight

T cos = mg -----(1)

T sin = qE -----------(2)

tan = qE/mg

tan 17.4 = 1.33*10^-6*E / 12.6*10^-3*9.8

E= 29095 N/C

so the magnitud of electric field is 2.9095*10^4 N/C is the correct answer