Question

#2) A charge q,-+200pC has . mass 0400 kg and is hung from the ceiling by a light string of length 1,-080 m. A second charge o,-150 JC is directly below the first charge, and is pulled upward by its atnraction to the first charge. It is connected to the floor by a string whose length is 1,-1.50 m. The height of the ceiling is (a) What is the electric force between the two charges? (b) What is the tension in the string connected to the ceiling? h 2.80 m IDENTIFY: What is the direction of the electric force on charge g, Upward )Downward PLAN: The charges are at rest and remaining at rest, and the net force on q, is-- )directed upward SOLVE: Solve the problem. )directed downward 2 INTERPRET: Al else being equal, which one of the following would NOT increase the tension in the string connected to the ceiling? () Decrease the height of the ceiling ) Increase the length of the string connected to the floor )Decrease the length of the string connected to the ceiling.

Answer 1

(a) Distance b/w charges = 2.8 - 1.5 - 0.8 = 0.5 m Electric force b/w 2 charges is given Vertical-bar FE Vertical-bar = Vertical-bar kq_1 q_2/r^2 Vertical-bar q_1 = 20 times 10^-6 c q_2 = -15 times 10^-6 c r = 0.5 Vertical-bar FE Vertical-bar = 1/4 pi epsilon_0 20 times 15 times 10^-12/(0.5)^2 = 10.79 N epsilon_0 = 8.85 times 10^-12 charges are of opposite sign so force is attractive q_1 attracted towards q_2 so force is downward on q_1 (b) FBD of q_1 T = Fe + mg = 10.79 + 0.4 times 9.81 = 14.714 As charge are on rest net force must be 0 Interpret: Decreasing weight increase Fe so increase in tension In correct Increase in length of any string decrease distance which increase Fe which increase Tension. In Correct Decreasing length decrease Fe which decrease tension \r\n This a question of conservation energy Initially potential energy PE_I = 1/4 pi epsilon_0 q_A q_B/r_AB r_AB = 0.5

2

This is a question of conservation of energy.