Question

A small 1.00-g plastic ball that has a charge q of 1.10 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 below. 1) If the ball is in equilibrium when the string makes a 9.80. angle with the vertical as indicated by , what is the electric field strength? (Express your answer to three significant figures.) mN/C Submit

Answer 1

T righarrow Tension theta = 9.89 degree m = 1.00 g = 1 times 10^- 3 kg q = 1.10 c g = 9.8 m/s^2 in equilibrium T sin theta = q E (1) T cos theta = mg (2) dividing (1) by (2) T sing theta/T cos theta = qE/mg implies qE/mg = tan theta implies E = mg tan theta/q = 1 times 10^-3 kg times 9.8 m/s^2/1.1c tan (9.80 degree) E = 9.8 times 10^- 3 N/1.1c times tan (9.80 degree) = 9.8 times 0 .173 times 10^- 3/1.1 N/C = 1.54 times 10^-3 N/C = 1.54 m N/c