Question

(a) Two insulated charged copper spheres A and B have their centresseparated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10–7 C? Theradii of A and B are negligible compared to the distance ofseparation. (b) What is the force of repulsion if each sphere is charged doublethe above amount, and the distance between them is halved?

Answer 1

a)

Given: The charge on sphere A is 6.5× 10 −7  C, charge on sphere B is 6.5× 10 −7  C and the distance between sphere A and sphere B is 50 cm.

The forces between two charged particle is given as,

F= 1 4π ε 0 ⋅ q A q B r 2 (1)

Where, q A is the charge on sphere A, q B is the charge on sphere B, r is the distance between charges.

By substituting the given values in the above equation, we get

F=9× 10 9 ⋅ ( 6.5× 10 −7 ) 2 0.5 2 =1.52× 10 −2  N

Thus, the force between two sphere is 1.52× 10 −2  N.

b)

When charge on each sphere is doubled and distance is becomes half, then equation (1) becomes,

F ′ = 1 4π ε 0 ⋅ ( 2 q A )( 2 q B ) ( 1 2 r ) 2 =16× 1 4π ε 0 ⋅ q A q B r 2 =16×F

By substituting the value of F in the above equation, we get

F ′ =16×1.52× 10 −2 =0.243 N

Thus, the new force between two spheres will be 0.243 N.