Question

A pendulum bob of mass 30.7x10 kg and carrying a charge 2x10 C is at rest in a horizontal
uniform electric field of 20000 V/m. The tension in the thread of the pendulum is (g =9.8m/s?)
(a) 3x10^-4 N (b) 4x10^-4 N
(C)5x10^-4N (d)6x10^-4N
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Answer 1

Fe = qE = 2×10^-8 × 20000

= 4× 10^-4

Tcosa = mg = 30.7× 10^-6 × 10 = 30.7× 10^-5

Tsin a = fe = 4× 10^-4

sin a= 4× 10^-4/T

T√1- sin^2 a = 30.7 × 10^-5

T√1-16×10^-8/T^2 = 30.7× 10^-5

T^2( 1 - 16× 10^-8/T^2) = 942.49× 10^-10

T^2 - 16× 10^-8 = 942.49 × 10^-10

T^2 = 10^-8( 16 + 942.49×10^-2)

= 10^-8( 16 + 9.42)

= 25.42 × 10^-8

So T ~5 × 10^-4

C)