Question

Two spheres a and b of radii 4 cm and 6 cm given charge of 80 microcoulomb and 20 microcoulomb respectively if these are joined by conducting wire then the charge flow between them will be

Answer 1

A = 4cm

B = 6cm

charge will flow from higher potential to lower potential.

•°• q1 = (80 - q)

q2 = (20 + q)

At time:

v1 = v2 ..(potential difference)

k(q1)/r1 = k(q2)/r2

k(80 - q)/4 = k( 20 + q)/6

(80 - q)/4 = (20 + q)/6

6/4 = (20 + q)/(80 - q)

3/2 = (20+ q) / (80 - q)

3(80 - q) = 2(20 + q)

240 - 3q = 40 + 2q

240 - 40 = 2q + 3q

200 = 5q

q = 200/5

q = 40

•°• q = 40µc charge will flow between them