QJEEM22Ph9a9c: JEE Main 2022 Physics - Class 12 JEEM22Ph9a9c Solution: Two identical metallic spheres and when placed at certain…

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

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

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 $= {q \over 2}$

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

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

Now, C is place midpoint between A and B.

Repulsion force between A and C,

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

Repulsion force between B and C,

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

$\therefore$ Net force on C,

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

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

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

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

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