Question

Two charged metal balls of a different size, radius R1 is bigger than R2, and then connected by a wire. Which ball gets the larger charge? What about charge density? Prove the answer.

Answer 1

charges will be more on the surface of the bigger ball and whose radius is R1 as R1 is bigger here.

higher the radius lower the charge density hence here R1 has lower charge density

here we have given that,

Two charged metal balls of a different size, and then connected by a wire.

radius R1 is bigger than R2

and then connected by a wire.

Now when the balls are connected by a wire, then defiantly the charge will move from one metal ball to another and when two metal balls are electrically connected, the charge distributes in such a way so that the potential of both objects is the same. Also charged ball’s potential is directly proportional to the charge of the ball and in reciprocal proportion to the ball’s radius.

mathematically we have relation as -

V=kQ/r

So the potential of the first ball is: V1=kQ1/R1

and the potential of the second ball is: V2=kQ2/R2

Since the potentials of both balls are the same.

V1 = V2

kQ1/R1 = kQ2/R2

here the constant k is the same in both the sides so that,

Q1/Q2=R1/R2 .........................1

so that form the above relation we can say that charges distributed in the same ratio as the ratio of balls radii.

hence we can say that there is more charge on the surface of the bigger ball and whose radius is R1 as R1 is bigger here.

now for the charge density we know that, the surface charge density on the ball depends on the total charge on the ball and the ball surface. so that

Q=σS

where

σ is the charge density

S surface area of the ball

and , σ1 = Q1/π R1^2

σ2 = Q2/π R2^2

so that,

σ1/σ2=Q1R2^2/Q1R1^2 .....................2

as form the equation 1 we have

Q1/Q2=R1/R2 so that equation 2 becomes now  

σ1/σ2 = R2/R1

so from here we can say that higher the radius lower the charge density hence here R1 has lower charge density