Q. 234.0( 6 Votes )

Two identical pith balls are charged by rubbing against each other. They are suspended from a horizontal rod through two strings of length20 cm each, the separation between the suspension points being 5 cm. In equilibrium, the separation between the balls is 3 cm. Find the mass of each ball and the tension in the strings. The charge on each ball has a magnitude 2.0 × 10–8 C.

Answer :


Length of strings,

Distance between suspension points,

Magnitude of charge on each ball,

Distance between the two balls,

Let the mass of each ball be m. Let the magnitude of Tension be T and the electric force between the balls be Fe.

To be in accordance with the fact, , the balls must be attracted to each other. Hence, they are oppositely charged.

By trigonometry,

At equilibrium, all forces must cancel out in accordance with newton’s first law. Hence,


Formula used:

By Coulomb’s law, the electric force is given by:

Where ϵ0 is the permittivity of free space

q1 and q2 are the magnitude of charges

r is the distance of separation between the charges

(Here, )


The magnitude of the electric force, Fe is given by:

Substituting values in (3), we get

From equation (1), we get

From equation (2), we get

Hence, mass of each of the balls is 8.2 grams and Tension in each of the ropes is 0.08N.

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