A conducting sphere S_1 of radius r_1 = 3 tex | vedclass

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(A) When two conducting spheres are connected by a wire,charge flows until their electric potentials become equal. Let $q_1$ and $q_2$ be the final ch

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{17}{6}$

Vedclass · Answer

(A) When two conducting spheres are connected by a wire,charge flows until their electric potentials become equal. Let $q_1$ and $q_2$ be the final charges on $S_1$ and $S_2$ respectively.By conservation of charge: $q_1 + q_2 = 10$.Since the potentials are equal: $\frac{k q_1}{r_1} = \frac{k q_2}{r_2} \Rightarrow \frac{q_1}{3} = \frac{q_2}{2} \Rightarrow q_1 = 1.5 q_2$.Substituting $q_1$ in the charge conservation equation: $1.5 q_2 + q_2 = 10 \Rightarrow 2.5 q_2 = 10 \Rightarrow q_2 = 4 	ext{ units}$.Thus,$q_1 = 6 	ext{ units}$.The potential $V$ at a point $P$ at distance $d_1 = 4 	ext{ cm}$ from the centre of $S_1$ and $d_2 = 3 	ext{ cm}$ from the centre of $S_2$ is the sum of potentials due to both spheres:$V = \frac{k q_1}{d_1} + \frac{k q_2}{d_2} = k \left( \frac{6}{4} + \frac{4}{3} ight) = k \left( \frac{3}{2} + \frac{4}{3} ight) = k \left( \frac{9 + 8}{6} ight) = k \frac{17}{6}$.Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we get $V = \frac{1}{4 \pi \varepsilon_0} \frac{17}{6}$.

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