Question

A 3.9 g aluminum foil ball with a charge of +4.6×10−9 C is suspended on a string in a uniform horizontal E⃗ field. The string deviates to the right and makes an angle of 30∘ with the vertical. Determine the magnitude of the electric field.

Answer 1

m = mass of the ball = 3.9 g = 0.0039 kg

q = charge on the ball = 4.6 x 10^-9 C

E = electric field in the region

F_e = electric force on the ball

T = tension force in the string

Electric force on the ball due to electric field is given as

F_e = qE

From the force diagram , force equation in vertical direction is given as

T Cos30 = mg                                                 eq-1

From the force diagram , force equation in horizontal direction is given as

T Sin30 = F_e                                               

T Sin30 = qE                                                          eq-2

Dividing eq-2 by eq-1

T Sin30/(T Cos30) = qE/(mg)

tan30 = qE/(mg)

tan30 = (4.6 x 10^-9)E/((0.0039)(9.8))

E = 4.8 x 10⁶ N/C