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(B) The potential $V$ at a distance $x$ from the center of a ring of radius $r$ carrying charge $q$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \

Vedclass · Accepted Answer

$-\frac{1}{5} \cdot \frac{q^{2}}{4 \pi \varepsilon_{0} r}$

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(B) The potential $V$ at a distance $x$ from the center of a ring of radius $r$ carrying charge $q$ is given by $V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{\sqrt{x^{2} + r^{2}}}$.For point $A$ at distance $x_A = \frac{4}{3}r$, the distance from the ring circumference is $d_A = \sqrt{(\frac{4}{3}r)^2 + r^2} = \sqrt{\frac{16}{9}r^2 + r^2} = \sqrt{\frac{25}{9}r^2} = \frac{5}{3}r$.Thus, $V_A = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{5r/3} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3q}{5r}$.For point $B$ at distance $x_B = \frac{3}{4}r$ on the opposite side, the distance from the ring circumference is $d_B = \sqrt{(\frac{3}{4}r)^2 + r^2} = \sqrt{\frac{9}{16}r^2 + r^2} = \sqrt{\frac{25}{16}r^2} = \frac{5}{4}r$.Thus, $V_B = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{5r/4} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4q}{5r}$.The work done $W$ in moving a charge $-q$ from $A$ to $B$ is $W = (-q)(V_B - V_A)$.$W = -q \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4q}{5r} - \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3q}{5r} \right) = -q \left( \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{5r} \right) = -\frac{1}{5} \cdot \frac{q^2}{4 \pi \varepsilon_{0} r}$.

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