Two similar conducting balls of mass 75 grams are hung from silk threads of leng

Question

Two similar conducting balls of mass 75 grams are hung from silk threads of length 165cm and carry equal but opposite charges of magnitude Q= 17.8

Two similar conducting balls of mass 75 grams are hung from silk threads of length 165 cm and carry equal but opposite charges of magnitude Q is 17.8 mu C. Additionally, a uniform electric field is applied to the left as shown. Determine the magnitude of the electric field that enables the balls to be in equilibrium (as shown) given theta = 25 degree.

Explanation / Answer

r = L*sin25 = 1.65*sin25 = 0.697 m

distance between two charges, d = 2*r = 1.395 m

Fe = k*Q^2/d^2 = (9*10^9*(17.8*10^-6)^2)/(1.395^2)= 1.465 N

as the charges are in equilibrium, the net force acting on the two charges in x and y direction are zero.

in x direction

E*Q - Fe - T*sin25 = 0

EQ - Fe = T*sin25

EQ - 1.465 = T*sin25

T*sin(25) = E*Q - 1.465.....(1)

in y direction

M*g = T*cos25

T*cos25 = 0.07589.81 = 0.73575 N.....(2)

1/2

tan25 = (E*Q - 1.465)/0.73575


E = (tan(25)*0.73575 + 1.465 )/Q

E = (tan(25)*0.73575 + 1.465 )/(17.8*10^-6)


E = 1.0158*10^5 N/C

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