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(B) The electric potential at the center $A$ of the first ring is the sum of the potential due to ring $1$ and ring $2$: $V_A = \frac{Q_1}{4\pi \varep

Vedclass · Accepted Answer

$\frac{q(Q_1 - Q_2)(\sqrt{2} - 1)}{4\pi \varepsilon_0 R\sqrt{2}}$

Vedclass · Answer

(B) The electric potential at the center $A$ of the first ring is the sum of the potential due to ring $1$ and ring $2$: $V_A = \frac{Q_1}{4\pi \varepsilon_0 R} + \frac{Q_2}{4\pi \varepsilon_0 \sqrt{R^2 + R^2}} = \frac{1}{4\pi \varepsilon_0 R} \left( Q_1 + \frac{Q_2}{\sqrt{2}} \right)$ The electric potential at the center $B$ of the second ring is: $V_B = \frac{Q_2}{4\pi \varepsilon_0 R} + \frac{Q_1}{4\pi \varepsilon_0 \sqrt{R^2 + R^2}} = \frac{1}{4\pi \varepsilon_0 R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} \right)$ The work done $W$ to move a charge $q$ from $A$ to $B$ is given by $W = q(V_B - V_A)$: $V_B - V_A = \frac{1}{4\pi \varepsilon_0 R} \left( Q_2 + \frac{Q_1}{\sqrt{2}} - Q_1 - \frac{Q_2}{\sqrt{2}} \right)$ $V_B - V_A = \frac{1}{4\pi \varepsilon_0 R} \left( Q_2(1 - \frac{1}{\sqrt{2}}) - Q_1(1 - \frac{1}{\sqrt{2}}) \right)$ $V_B - V_A = \frac{1}{4\pi \varepsilon_0 R} (Q_2 - Q_1) \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right)$ Since $W = q(V_B - V_A)$,we have: $W = \frac{q(Q_2 - Q_1)(\sqrt{2} - 1)}{4\pi \varepsilon_0 R\sqrt{2}}$ Note: The magnitude of work is $\frac{q(Q_1 - Q_2)(\sqrt{2} - 1)}{4\pi \varepsilon_0 R\sqrt{2}}$.

Two rings of radius $R$ are placed coaxially at a distance $R$ apart. They carry charges $Q_1$ and $Q_2$ respectively. How much work is required to move a charge $q$ from the center of one ring to the center of the other?

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