Question

A small 1.45 g plastic ball that has a charge q of 1.75 C is suspended by a string that has a length L of 1.00 m in a uniform electric field, as shown in the figure. If the ball is in equilibrium when the string makes a 9.80 angle with the vertical as indicated by θ, what is the electric field strength E?

Answer 1

here,

mass , m = 1.45 g = 0.00145 kg

charge , q = 1.75 C

L = 1 m

theta = 9.8 degree

let the tension in the string be T and electric feild be E

equating the forces vertically

T * cos(theta) = m * g ....(1)

and

equating the forces horizontally

T * sin(theta) = q * E ....(2)

from (1) and (2)

tan(theta) = ( q * E / (m * g))

tan(9.8) = ( 1.75 * E /( 0.00145 * 9.81))

solving for E

E = 1.4 * 10^-3 N/C

the magnitude of electric field is 1.4 * 10^-3 N/C