Question

A large positively charged object with charge q + = 3.75 μC is brought near a negatively charged plastic ball suspended from a string of negligible mass. The suspended ball has a charge of q − = − 53.3 nC and a mass of 12.5 g. What is the angle the string makes with the vertical when the positively charged object is 22.5 cm from the suspended ball? The positively charged object is at the same height as the suspended ball.

Answer 1

Solution : Electric - force Fe = 9. on ita Fe = 2 K 2 2 2 viernes 9 T. q- Fe = 13.75*10*6C) (8.99x109 Nm /c) (53.3 X10-90) (22.5 x 10-2 m) 2 ; DE Fe = 3.549x10-N. Weight of 9. : W = mg = (12.5 810**kq)(9.81m) W = 12.26 X 10-2N. At q. is in equilibrium : fx=0. EF, 0 q+E - Tsino = 0 , Troso - mg =0. ( 3.549x10*N) = Tsino-0 Teos = 112.26x1o N) Dividing ① by ②, we get, T sino 3.549x10²N 12.26 X10-2 N Toso 0 -16.1450

The angle the string makes with the vertical is 16.145 ^o