First slide

Electrostatic force

Question

Two identical conducting spheres having unequal positive charges q₁ and q₂ separated by distance r. If they are made to touch each other and then separated again to the same distance. The electrostatic force between the spheres in this case will be (neglect induction of charges)

Easy

A

less than before
B

same as before
C

more than before
D

zero

Solution

When spheres are made to touch each other, charges distribute equally. Final charge on both sphere is $\frac{q_{1} + q_{2}}{2}$
$\begin{array}{l} I n i t i a l l y : F_{1} \propto q_{1} q_{2} \\ F i n a l l y : F_{2} \propto \frac{{(q_{1} + q_{2})}^{2}}{4} \\ \Rightarrow F_{2} \propto \frac{{q_{1}}^{2}}{4} + \frac{{q_{2}}^{2}}{4} + \frac{q_{1} q_{2}}{2} \\ Thus, F_{2} > F_{1} \end{array}$

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