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Get Answers to all your Questions
Q.14) Two identical charged conducting spheres A and B have their centers separated by a certain distance. The charge on each sphere is $q$, and the force of gepulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere $A$ first and then with $B$ and finally removed from both. New force of repulsion between spheres A and B (Radii of A and B are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :
A) $\frac{3 F}{8}$
B) $\frac{3 F}{5}$
C) $\frac{2 F}{3}$
D) $\frac{F}{2}$
Answers (1)
Solution:
Initial force between A and B :
$$
F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r^2}
$$
After a third identical uncharged sphere is successively touched with $A$ and then $B$, charge redistributes:
After contact with A: charge on each becomes $q / 2$
Then contact with $B$ : total charge $=q / 2+q=3 q / 2 \rightarrow$ each gets $3 q / 4$ So final charges:
$A=q / 2$
$B=3 q / 4$
New force:
$$
F^{\prime}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{(q / 2)(3 q / 4)}{r^2}=\frac{3 q^2}{8 r^2}
$$
Divide by original force:
$$
\frac{F^{\prime}}{F}=\frac{3 q^2 / 8 r^2}{q^2 / r^2}=\frac{3}{8}
$$
Hence, the answer is option (1) $3 F / 8$.
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