Question

A small object has charge Q. Charge q is removed from it and placed on a second small object The two objects are placed I m apart. For the force that each object exerts on the other to be a maximum, what should q be? a) q 20 c) q Q/2 d) q /4

Answer 1

Answer:

q = Q/2

option c is correct

Explanation:

given data

charge = Q

distance r = 1 m

charge = q

to find out

what should q be

solution

we have given charge q is remove so

q1 = Q-q

q2 = q

we know by coulombs law force

force = kq1q2 / r²   .............1

put here value and we know k is constant

F = k (Q-q) q / 1

now take derivative charge q and put = 0

dF/dq = k d(Q-q)q / dq

so  k (Q-q)q =0

Q = 2q

q = Q/2

so option c is correct