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

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(B) Let the centres of the rings be $A$ and $B$. The potential at the centre of a ring of radius $R$ with charge $Q$ is $V = \frac{1}{4\pi \varepsilon

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the centres of the rings be $A$ and $B$. The potential at the centre of a ring of radius $R$ with charge $Q$ is $V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R}$.The potential at centre $A$ due to ring $A$ (charge $+q$) is $V_{A1} = \frac{1}{4\pi \varepsilon_0} \frac{q}{R}$.The potential at centre $A$ due to ring $B$ (charge $-q$) is $V_{A2} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{\sqrt{R^2 + d^2}}$.So,$V_A = V_{A1} + V_{A2} = \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} ight]$.The potential at centre $B$ due to ring $B$ (charge $-q$) is $V_{B1} = \frac{1}{4\pi \varepsilon_0} \frac{-q}{R}$.The potential at centre $B$ due to ring $A$ (charge $+q$) is $V_{B2} = \frac{1}{4\pi \varepsilon_0} \frac{q}{\sqrt{R^2 + d^2}}$.So,$V_B = V_{B1} + V_{B2} = \frac{q}{4\pi \varepsilon_0} \left[ -\frac{1}{R} + \frac{1}{\sqrt{R^2 + d^2}} ight] = -V_A$.The potential difference between the centres is $\Delta V = V_A - V_B = V_A - (-V_A) = 2V_A$.$\Delta V = 2 	imes \frac{q}{4\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} ight] = \frac{q}{2\pi \varepsilon_0} \left[ \frac{1}{R} - \frac{1}{\sqrt{R^2 + d^2}} ight]$.

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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