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1. Electric Charges and Fields
hard
Two equally charged, identical metal spheres $A$ and $B$ repel each other with a force '$F$'. The spheres are kept fixed with a distance '$r$' between them. A third identical, but uncharged sphere $C$ is brought in contact with $A$ and then placed at the mid-point of the line joining $A$ and $B$. The magnitude of the net electric force on $C$ is
A
$F$
B
$3F/4$
C
$F/2$
D
$F/4$
Solution
(a) $F = k\frac{{{Q^2}}}{{{r^2}}}$
Force on $C$ due to $A$, ${F_A} = \frac{{k{{(Q/2)}^2}}}{{{{(r/2)}^2}}} = \frac{{k{Q^2}}}{{{r^2}}}$
Force on $C$ due to $B$, ${F_B} = \frac{{KQ(Q/2)}}{{{{(r/2)}^2}}} = \frac{{2K{Q^2}}}{{{r^2}}}$
Net force on $C$, ${F_{net}} = {F_B} – {F_A} = \frac{{k{Q^2}}}{{{r^2}}} = F$
Standard 12
Physics
