Question

Two small metallic spheres, each of mass 1.0 g are suspended as pendulums by light strings from a common point. The spheres are given the same electric charge and it is found that they come to equilibrium when each string is at an angle of 10.0 degrees with the vertical. If each string is 1.0 m long, what is the magnitude of the charge on each sphere?

Answer 1

Answer:

Q = 1.52 X 10⁻⁷ C.

Explanation:

String is inclined at 10 degree with the vertical so the vertical component of T will balance the weight and the horizontal component will balance the force of repulsion F.

T Cos 10 = mg

T Sin 10 = F

Tan 10 = F / mg

F = mg Tan 10

= 10⁻³ x 9.8 x .176

= 1.728 x 10⁻³ .

Force of repulsion F = K Q² / R²

Distance between two charges = 2 x 1 x sin 10

= .347 m

1.728 X 10⁻³ = $[tex]\frac{9\times10^9\times Q^2}{(0.347)^2}$[/tex]

Q² = 2.31 X 10⁻¹⁴

Q = 1.52 X 10⁻⁷ C.