Question

Repulsive force F is acting between two identical charged sphere. Now, another charged sphere C is first kept nearer to A then, separated now, it is kept at midpoint of A & B. Calculate net force on it.​

Answer 1

Answer:

When a third sphere is brought in contact with sphere B, the charges moves to the new sphere so that their potentials are equal. This distributes charges in half between the both spheres. Now when it is kept in between A and B, it experiences a force calculated as follows:

F=kq2r2F=kq2r2 (initial force between A and B)

On c, force due to A is : kq^2/(2*(r/2)^2) = 2kq^2/r2 = 2F towards B

Force due to B, kq^2/r^2 = F . So net force is 2F - F = F towards sphere A.

Answer 2

Answer

(i) Charge on C is same as A and B, Force = 0

(ii) Charge on C is opposite but same as A and B in magnitude, Force = 0

Explanation:

In the question, it isn't mentioned whether C is identical with A and B or not. So, let us assume all 3 to be identical.

Initially, A and B spheres are kept at a distance d

So, the Force between them becomes

$F = \frac{kq_1q_2}{d^2} ; \ where \ k = \frac{1}{4\pi \epsilon_0}$ ; q₁ and q₂ are the charges on A and B

Now, C is kept near A, then the charged on both of them will be distributed. As charge of C is not mentioned, we will do in 2 methods (i) Charge on C is  identical to A and B

So, when they are kept nearer to each other, then no change will be transferred between them as they r identical

Now, We keep C between A and B,

Now the force on C will be like, towards its left will be the Force dur to B and towards its right will be the force due to A. And these 2 forces are identical, as charges and distances are same. Hence, both the forces cancel out to give out force '0' on sphere C

So, (i) Force on sphere C = 0

(ii) The second method is, charge on C is -of opposite but identical to A and B in magnitude

So, when we bring C nearer to A, as they both are identical, Charges on A and C will be zero. Now Charge on A and C = 0

Now, keeping C in between of A and B, there will be no force on C due to A as A is uncharged sphere now. Force on C will be only due to B. So,

Force on C due to B is given by

$F' = \frac{kq(0)}{(\frac{d}{2} )^2} = 0$ [ Charge on C is 0]

So, in the both cases, the answer would be 0

$\boxed {\sf{NOTE : }}$ Your question would be valid, if "A and B are two identical spherical charged bodies which repel each other with force F, kept at a finite distance apart A third uncharged sphere of the same size is brought in contact with sphere B and removed. It is then kept at mid-point of A and B find the magnitude of force on C"

• The answer of this question is F
• Explanation :

A and B are two identical spherical charged bodies which repel each other with force =F

finite distance = r between them

As C is uncharged, if it is brought nearer to A, the charge on A and C will be q/2

Now, Charges on A and C = q/2 and B = q

Let $F_{CB}$ be the force on C due to B and $\ F_{CA}$ be the force on C by A

Now,

$F_{CA} = \frac{k (\frac{Q}{2} )(\frac{Q}{2} )}{(\frac{r}{2}^2 )} = \frac{kq^2}{r^2} = F\\\\F_{CB} = \frac{k(\frac{q}{2} )(q)}{(\frac{r}{2} )^2} = \frac{2kq^2}{r^2} = 2F$

As these both are opposite, Resultant is 2F - F = F

Hope it helps!!

{I gave a explanation in ur another account[SorkoZom](in one math question), please read it.}