Question

Objects A,B and C are three insulated, spherical conductors of radii R, 2R and 3R respectively. Originally, A and B have charges of +3 mc whereas C has a charge of -6 mc. Objects A and C are touched and moved apart. Calculate the new charges appears over A, B and C.​

Answer 1

Answer:

A: $\\\tt \dfrac{-3mC}{10}$

C: $\\\tt \dfrac{-27mC}{10}$

B: $\\\tt +3mC$

Explanation:

Let the the area for charge distributions in each of the speheres,

For sphere A area = $\\\tt \pi R^2$

For sphere B it is $\\\tt 4 \pi R^2$

and for sphere C it's area = $\\\tt 9\pi R^2$

Now when A and C are touched the charges will enforce force upon each of other charges, so that the net charge is of -3mC (on area of A and C which is

$\\\tt \pi R^2+ \pi R^2= 10 \pi R^2$)

Before dividing the charge into A and C spheres you should know that -3mC charge holding particles are of same type and hence they will gain equilibrium only when each of them is as far as possible from other charges.

So we assume that once the charges have gained equillibrium change per unit area wil be (during the time when the metal balls are bing touched),

$\\\tt \dfrac{-3mC}{10\pi R^2}$

Therefore new charge on A after separation ( of area $\\\tt \pi R^2$) will be $\\\tt \dfrac{-3mC}{10}$

and charge on C (of area $\\\tt9\pi R^2$) will be $\\\tt\dfrac{-27mC}{10}$

Since there was no interference with B there will be no change in its charge.