Question

Two metal spheres of identical mass m = 4.20 g are suspended by light strings 0.500 m in length. The left-hand sphere carries a charge of 0.785 µC, and the right-hand sphere carries a charge of 1.47 µC. What is the equilibrium separation between the centers of the two spheres?

Answer 1

Step-by-step explanation:

Answer:

0.632 m

Step-by-step explanation:

let the equilibrium separation between the charges is d and the angle made by string with the vertical is θ.

According to the diagram,

d = L Sinθ + L Sinθ = 2 L Sinθ .....(1)

Let T be the tension in the string.

Resolve the components of T.

T Sinθ = k q1 q2 / d^2

T Sinθ = k q1 q2 / (2LSinθ)² .....(2)

T Cosθ = mg .....(3)

Dividing equation (2) by equation (3), we get

tanθ = k q1 q2 / (4 L² Sin²θ x mg)

tan θ Sin²θ = k q1 q2 / (4 L² m g)

For small value of θ, tan θ = Sin θ

So,

Sin³θ = k q1 q2 / (4 L² m g)

Sin³θ = (9 x 10^9 x 0.785 x 10^-6 x 1.47 x 10^-6) / (4 x 0.5 x 0.5 x 4.20 x 10^-3 x 9.8)

Sin³θ = 0.2523

Sinθ = 0.632

θ = 39.2 degree

So, the separation between the two charges, d = 2 x L x Sin θ

d = 2 x 0.5 x 0.632 = 0.632 m