Question

Two small identical spheres each of mass 1g and carry 10to the power -9c are suspended by thread of equal length. distance between sphere is 0.3 in equilibrium then the inclination of thread with vertical will be​

Answer 1

see free body diagram,

Let Θ is angle made by thread with the vertical.

at equilibrium,

upward force = downward force

TcosΘ = mg ....(1)

forward force = backward force

TsinΘ = Fe = kq²/r² .....(2)

from equations (1) and (2),

TsinΘ/TcosΘ = kq²/r²mg

or, tanΘ = kq²/r²mg

here, k = 9 × 10^9, q = 10^-9C, r = 0.3m

m = 1g = 10^-3 kg

so, tanΘ = 9 × 10^9 × (10^-9)²/{(0.3)² × 10^-3 × 10}

= 9 × 10^-9/(0.09 × 10^-2}

= 10^-5

so, Θ = tan^-1(10^-5)

hence, angle made by thread with the vertical is tan^-1(10^-5)

Answer 2

Explanation:

hope it will help you friends