Question

64 drops of radius 0.02m and each carrying a charge of 5 microcoulmb are combined to form a bigger drop . Find how the surface charge density of electrification will change if no charge is lost.​

Answer 1

Answer:

See the image

Explanation:

Hope you understand

Answer 2

Answer:

1:4

Explanation:

volume of each small drop= 4/3π(0.02)^3 m^3

volume of 64 small drops= 4/3π(0.02)^3 × 64 m^3

let R be the radius of bigger drop formed.

Then, 4/3πR^3= 4/3π(0.02)^3 × 64

R^3 = (0.02)^3× 4^3

therefore, R= 0.02×4= 0.08m

charge on small drop = 5µC = 5×10^-6C

surface charge density of small drop = q/4πr^2= 5×10^-6/4π(0.02)^2 Cm^-2

surface charge density of bigger drop= 5×10^-6 × 64/4π(0.08)^2 Cm^-2

therefore, surface charge density of small drop/surface charge density of bigger drop= 1/4= 1:4