Question

two small metal spheres are charged so that they repel each other with a force of 2 into 10 power minus 5 newton the charge on one sphere is twice that on other when they are removed 0.1 for the apart the force is reduced to 5 into 10 power minus 5 newton what are the charge and what is the initial separation between them? solve only images

Answer 1

Explanation:

Given Two small metal spheres are charged so that they repel each other with a force of 2 x 10-3 N. The charge on one sphere is twice that on other. When they are moved 0.1 m further apart the force reduces to 5 x 10-4 N. what are the charges and initial separation between them?

Given F = 2 x 10^-3 N

Let Q1 = q and Q2 = 2q

Let r be the distance between them.

We know that F = k Q1Q2 / r^2

So 2 x 10^-3 = k (q) (2q) / r^2

Or 2 x 10^-3 = 2 q^2 k / r^2

Or kq^2 / r^2 = 10^-3 -------------------1

Now r changes to (r + 0.1 )m then force will be 5 x 10^-4 N

So 5 x 10^-4 N = (k(q)(2q) / (r + 0.1)^2

So 5 x 10^-4 = 2q^2k / (r + 0.1)^2

From eqn 1 we get

So q^2k = r^2 x 10 ^-3

Therefore 5 x 10^-4 = 2 x 10^-3 x r^2 / (r + 0.1)^2

Therefore 2.5 x 10^-1 = (r / (r + 0.1))^2

So 0.25 = (r/(r + 0.1))^2

0.5 = r / (r + 0.1)

0.5 r + 0.05 = r

So r = 0.1 m

So from 1 we get 9 x 10^9 x q^2 / (0.1)^2 = 10^-3

So q^2 = 10^-3 x 10^-2 / 9 x 10^9 = 10^- 14 / 9

So q = 10^-7 / 3

So q = 3.33 x 10^-8 C

So 2q = (2/3) x 10^-7

= 6.66 x 10^-8 C

Therefore charges are 33.33 x 10^-9 C and 66.66 x 10^-9 C