Question

A spherical drop of water has 3 x 10⁻¹⁰ C amount of charge residing on it. 500 V electric potential exists on its surface. Calculate the radius of this drop. If eight such drops (Having identical charge and radii) combine to form a single drop, calculate the electric potential on the surface of the new drop. k = 9 x 10⁹ SI.

Answer 1

Dear Student,

◆ Answer -

r = 5.4 mm

V' = 2000 V

● Explanation -

Let r be the radius of the drop. Electric potential of the drop is given by -

V = (1/4πε) q/r

r = (1/4πε) q/V

r = 9×10^9 × 3×10^-10 / 500

r = 5.4×10^-3 m

When 8 such drops are merged together, radius of bigger drop will be -

R = r.∛n

R = 5.4×10^-3 × ∛8

R = 5.4×10^-3 × 2

R = 1.08×10^-2 m

Total charge on bigger drop will be -

Q = nq

Q = 8 × 3×10^-10

Q = 2.4×10^-9 C

Now, electric potential of bigger drop will be -

V' = (1/4πε) Q/R

V' = 9×10^9 × 2.4×10^-9 / 1.08×10^-2

V' = 2000 V

Hope this helps you...