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Work Done When Force Is Varying

Two identical...

Question

Two identical spheres having charges $+ q_{1}$ and $+ q_{2}$ are kept at a distance "d". These two are brought in contact with an insulated handle and again separated to the same distance. Find the force $F_{2}$ between them in terms of initial force $F_{1}$ .

A

F_{2} = F_{1} \frac{(q_{1} + q_{2})^{3}}{4 q_{1}^{2} q_{2}}

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B

F_{2} = F_{1} \frac{(q_{1} + q_{2})^{3}}{4 q_{1} q_{2}^{2}}

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C

F_{2} = F_{1} \frac{(q_{1} + q_{2})}{4 q_{1}}

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D

F_{2} = F_{1} \frac{(q_{1} + q_{2})^{2}}{4 q_{1} q_{2}}

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Solution

The correct option is D $F_{2} = F_{1} \frac{(q_{1} + q_{2})^{2}}{4 q_{1} q_{2}}$
Let initially, charges are separated by a distance $^{'} d^{'} .$
Then, according to Coulomb's Law,
$F_{1} = \frac{k q_{1} q_{2}}{d^{2}}$
$\Rightarrow d^{2} = \frac{k q_{1} q_{2}}{F_{1}}$ ...(i)

Now, when both the identical spheres are brought in contact, then charge on each sphere is average of initial charge on each sphere.
i.e, $Q = \frac{q_{1} + q_{2}}{2}$ on each sphere.

Now, Force between them i.e, $F_{2}$ is
$F_{2} = \frac{k Q^{2}}{d^{2}}$
From (i), $d^{2} = \frac{k q_{1} q_{2}}{F_{1}}$
So, $F_{2} = \frac{k (q_{1} + q_{2})^{2}}{k q_{1} q_{2} \times 4} \times F_{1}$
$F_{2} = F_{1} \times \frac{(q_{1} + q_{2})^{2}}{4 q_{1} q_{2}}$

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