Two identical thin rings,each of radius R tex | vedclass

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Question

(B) The work done in moving a charge $q$ from point $A$ to point $B$ is given by $W = q(V_B - V_A)$.The potential at the centre $A$ of the first ring

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The work done in moving a charge $q$ from point $A$ to point $B$ is given by $W = q(V_B - V_A)$.The potential at the centre $A$ of the first ring due to both rings is:$V_A = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q_1}{R} + \frac{Q_2}{\sqrt{R^2 + R^2}} ight) = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q_1}{R} + \frac{Q_2}{\sqrt{2}R} ight)$.The potential at the centre $B$ of the second ring due to both rings is:$V_B = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q_2}{R} + \frac{Q_1}{\sqrt{R^2 + R^2}} ight) = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q_2}{R} + \frac{Q_1}{\sqrt{2}R} ight)$.The work done is $W = q(V_B - V_A)$:$W = \frac{q}{4\pi \varepsilon_0} \left[ \left( \frac{Q_2}{R} + \frac{Q_1}{\sqrt{2}R} ight) - \left( \frac{Q_1}{R} + \frac{Q_2}{\sqrt{2}R} ight) ight]$.$W = \frac{q}{4\pi \varepsilon_0 R} \left[ (Q_2 - Q_1) + \frac{Q_1 - Q_2}{\sqrt{2}} ight]$.$W = \frac{q(Q_2 - Q_1)}{4\pi \varepsilon_0 R} \left( 1 - \frac{1}{\sqrt{2}} ight) = \frac{q(Q_2 - Q_1)}{4\pi \varepsilon_0 R} \left( \frac{\sqrt{2} - 1}{\sqrt{2}} ight)$.Since the question asks for work done moving from $A$ to $B$,and the options are expressed in terms of $(Q_1 - Q_2)$,we note that $W_{A 	o B} = -W_{B 	o A}$. The magnitude or the expression matching the options is $q(Q_1 - Q_2)(\sqrt{2} - 1) / (\sqrt{2} \cdot 4\pi \varepsilon_0 R)$.

Two identical thin rings,each of radius $R \text{ m}$,are coaxially placed at a distance $R \text{ m}$ apart. If $Q_1$ and $Q_2$ coulombs are the charges uniformly spread on the two rings respectively,the work done in moving a charge $q$ from the centre of one ring to that of the other is

Similar Questions

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