Question

two equally charged identical small balls kept fixed distance apart exert a repulsive force F on each other . A similar uncharged ball , after toucng one of them is placed at the mid point of the line joining the two balls . what will be the force experienced by the trd ball

Answer 1

Let Q charged on two identical balls and separation between them is 2R
Then, repulsive force act between them , F = KQ²/(2R)² = KQ²/4R² ----(1)

Now, a 3rd ball is touched with one of them{ assume 1 St ball } ,
Then, charge will share equally to both the balls.
e.g., charge on 1st ball = Q/2
charge on 3rd ball = Q/2.

Now, 3rd ball is placed midpoint of 1st and 2nd ball .
so, Force act on 3rd ball due to 1st ball , F₁ = K(Q/2)(Q/2)/R² = KQ²/4R²
again, force act on 3rd ball due to 2nd ball , F₂ = K(Q/2)(Q)/R² = KQ²/2R²
Because both the forces are repulsive nature and F₂ > F₁
so, net force experienced by 3rd ball = F₂ - F₁ = KQ²/2R² - KQ²/4R² = KQ²/4R²
Now, from equation (1)
Net force experienced by 3rd ball = F { e.g., force experienced by 1st and 2nd ball }

Answer 2

Answer:F

Explanation:

Let us assume there are two charges Q and Q placed at a distance of 2r

The repulsive force acting on it will be $F=\frac{KQ^{2} }{2r^{2} } =\frac{KQ^{2} }{4r^{2} }$    (eqn 1)  

After touching ball 3 with ball 1,the charge will be equally distributed

charge on first ball=Q/2

Charge on third ball=Q/2

$F1=\frac{K*Q/2*Q/2 }{r^{2} }$   (force on ball 1 due to 3) =$\frac{KQ^{2} }{4r^{2} }$

$F2=\frac{K*Q/2*Q}{r^{2} } =\frac{KQ^{2} }{2r^{2} }$ (force on ball 2 due to 3)

F3=F2-F1=$\frac{KQ^{2} }{2r^{2} } =\frac{KQ^{2} }{4r^{2} }$   (as F2>F1 )

as $\frac{KQ^{2} }{4r^{2} }= F$     (acc to eqn 1 )

Hence force on third ball will be=F