Two similarly and equally charged identical metal spheres A and B repel each other with a force of $2 \times 10^{-5}\,\text{N}$. A third identical uncharged sphere C is touched with A and then placed at the mid-point between A and B. Calculate the net electric force on C.

$2 \times 10^{-5}\,\text{N}$ (directed towards B).

CBSE Class 12 · 2003 · ElectrostaticsShort Answer

Two similarly and equally charged identical metal spheres A and B repel each other with a force of

2 \times 1 0^{- 5} N

. A third identical uncharged sphere C is touched with A and then placed at the mid-point between A and B. Calculate the net electric force on C.

Answer: $2 \times 1 0^{- 5} N$ (directed towards B).

Let each of A and B carry charge

Q

at separation

r

, so

F = \frac{k Q ^{2}}{r ^{2}} = 2 \times 1 0^{- 5} N

. Touching C (uncharged) to A shares the charge equally: A and C each become

\frac{Q}{2}

. Sphere C is placed at the mid-point, i.e. at

\frac{r}{2}

from both A and B.
Force on C due to A:

F_{A} = \frac{k ( Q /2 ) ( Q /2 )}{( r /2 ) ^{2}} = \frac{k Q ^{2}}{r ^{2}}

, directed towards B. Force on C due to B:

F_{B} = \frac{k ( Q /2 ) ( Q )}{( r /2 ) ^{2}} = \frac{2 k Q ^{2}}{r ^{2}}

, directed towards A.
Net force

= F_{B} - F_{A} = \frac{2 k Q ^{2}}{r ^{2}} - \frac{k Q ^{2}}{r ^{2}} = \frac{k Q ^{2}}{r ^{2}} = 2 \times 1 0^{- 5} N

, directed towards B.

CBSE Class 12 Board · 2003 · Subjective