JEE Main 2022 — Electrostatics Question with Solution

Let two identical spheres have charge q. And distance between them = r

$$\therefore$$ Force between the spheres $$(F)=\frac{k{q}^2}{r^2}$$

Now when an identical uncharged sphere C comes in contact with A, charge q on sphere A get's divided equally to both sphere.

So, both sphere A and C have charge $$=\frac{q}{2}$$

Now, C get's in contact with B. So their total charge $$\left(q+\frac{q}{2}\right)$$ gets divided equally.

So, charge on both B and C is $$=\frac{q+\frac{q}{2}}{2}=\frac{3q}{4}$$

Now, C is place midpoint between A and B.

Repulsion force between A and C,

$$F_{AC}=\frac{k\left(\frac{q}{2}\right)\left(\frac{3q}{4}\right)}{{\left(\frac{r}{2}\right)}^2}=\frac{4k\times 3q^2}{8r^2}$$

Repulsion force between B and C,

$$F_{BC}=\frac{k\left(\frac{3q}{4}\right)\left(\frac{3q}{4}\right)}{{\left(\frac{r}{2}\right)}^2}=\frac{4k\times 9q^2}{16r^2}$$

$$\therefore$$ Net force on C,

$$F_{net}=F_{BC}-F_{AC}$$

$$=\frac{4k\times 9q^2}{16r^2}-\frac{4k\times 3q^2}{8r^2}$$

$$=\frac{k{q}^2}{r^2}\left(\frac{9}{4}-\frac{3}{2}\right)$$

$$=\frac{k{q}^2}{r^2}\times \frac{3}{4}$$

$$=\frac{3}{4}F$$ [as $$F=\frac{k{q}^2}{r^2}$$]