Two thin wire rings each having a radius R ar | vedclass

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(D) Let the centres of the two rings be $O_1$ and $O_2$. The potential at any point on the axis of a charged ring of radius $R$ and charge $Q$ at a di

Vedclass · Accepted Answer

$\frac{q}{2\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$

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(D) Let the centres of the two rings be $O_1$ and $O_2$. The potential at any point on the axis of a charged ring of radius $R$ and charge $Q$ at a distance $x$ from its centre is given by $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{\sqrt{R^2 + x^2}}$.For the first ring (charge $+q$) at centre $O_1$:Potential due to itself is $V_{1, self} = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}$.Potential due to the second ring (charge $-q$) at distance $d$ is $V_{1, other} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + d^2}}$.So,$V_{O_1} = \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$.For the second ring (charge $-q$) at centre $O_2$:Potential due to itself is $V_{2, self} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{R}$.Potential due to the first ring (charge $+q$) at distance $d$ is $V_{2, other} = \frac{1}{4\pi \varepsilon_0} \frac{q}{\sqrt{R^2 + d^2}}$.So,$V_{O_2} = \frac{q}{4\pi \varepsilon_0} \left[ -\frac{1}{R} + \frac{1}{\sqrt{R^2 + d^2}} \right] = -V_{O_1}$.The potential difference is $\Delta V = V_{O_1} - V_{O_2} = V_{O_1} - (-V_{O_1}) = 2V_{O_1}$.$\Delta V = 2 \cdot \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right] = \frac{q}{2\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} \right]$.

Two thin wire rings each having a radius $R$ are placed at a distance $d$ apart with their axes coinciding. The charges on the two rings are $+q$ and $-q$. The potential difference between the centres of the two rings is

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