Question

Force between two charges is 36× 10^-6 N, when separated by a certain distance. On increasing the separation by 5m the force reduces to 25× 10- 6 N. If one charge is 10 times the other, find the initial separation and magnitude of charges.

Answer 1

Dear friend,

◆ Answer -

r = 25 m

q1 = 10^-7 C

q2 = 10^-8 C

● Explaination -

Let q1 & q2 be two charges such that q1 = 10 q1.

Initially,

F = k.q1.q2/r^2

36×10^-6 = k.q1.10q1/r^2

k (q1/r)^2 = 3.6×10^-6

k (q1)^2 = 3.6×10^-6 r^2 ...(1)

Later,

F' = k.q1.q2/r'^2

25×10^-6 = k.q1.10q1/(r+5)^2

k (q1/r+5)^2 = 2.5×10^-6

k (q1)^2 = 2.5×10^-6 (r+5)^2 ...(2)

Equating (1) & (2),

3.6×10^-6 r^2 = 2.5×10^-6 (r+5)^2

36 r^2 = 25 (r+5)^2

6r = 5 (r + 5)

r = 25 m

Putting r = 25 in (1),

k (q1)^2 = 3.6×10^-6 r^2

9×10^9 × q1^2 = 3.6×10^-6 × 25^2

q1^2 = (6×10^-3 × 5)^2 / 3×10^5

q1 = 10^-7 C

Putting this in q1 = 10q2,

q2 = q1 / 10

q2 = 10^-7 / 10

q2 = 10^-8 C

Thanks for asking...