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# A conducting sphere of radius $R$, and carrying a charge $q$ joined to a conducting sphere of radius $2R$, and carrying charge $-2q$. The charge flowing between them will beA. $q/3$B. $2q/3$C. $q$D. $4q/3$

Last updated date: 17th Sep 2024
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Hint:After the spheres are joined, basically the potential of the spheres will become the same but the total charge would be the same as before. So, we get two equations- one from the total charge and the other from the equal potential – on solving which we can get the amount of charge flow.

Consider the two spheres ${S_1}$ and ${S_2}$ with radii ${R_1},{R_2}$ and charges ${Q_1},{Q_2}$ respectively.
According to the question, we are given
${R_1} = R \\ \Rightarrow{R_2} = 2R \\ \Rightarrow{Q_1} = q \\ \Rightarrow{Q_2} = - 2q \\$
So, the total initial charge can be given by,
${Q_{ini}} = {Q_1} + {Q_2} = q + \left( { - 2q} \right) \\ \Rightarrow{Q_{ini}} = - q \\$
Now, on joining the two spheres the charge starts flowing from one sphere to another until both the spheres reach the same potential, but the total charge remains the same.
Let ${Q_1}^\prime ,{Q_2}^\prime$ be the new charges on spheres respectively and let their potentials be ${V_1},{V_2}$ after being joined. So, according to the conditions,
${Q_1}^\prime + {Q_2}^\prime = {Q_{ini}} \\ \Rightarrow {Q_1}^\prime + {Q_2}^\prime = - q \\$
Also, the potentials of both the spheres will be equal,
${V_1} = {V_2} \\ \Rightarrow \dfrac{{k{Q_1}^\prime }}{{{R_1}}} = \dfrac{{k{Q_2}^\prime }}{{{R_2}}} \\ \Rightarrow \dfrac{{{Q_1}^\prime }}{R} = \dfrac{{{Q_2}^\prime }}{{2R}} \\ \Rightarrow {Q_1}^\prime = \dfrac{{{Q_2}^\prime }}{2} \\$
On solving the above two equations we get,
${Q_2}^\prime = \dfrac{{ - 2q}}{3}$ and ${Q_1}^\prime = \dfrac{{ - q}}{3}$
Now, we can find the charge flow by subtracting the final charge from the initial charge on any of the spheres as
$\therefore{Q_{flow}} = q - \left( { - \dfrac{{2q}}{3}} \right) = \dfrac{{4q}}{3}$

Therefore, the correct option is D.

Note:We can find the charge flow from any of the spheres because the flow would be the same for both the spheres. Also, the main points to be noted here are that the total charge remains the same and the flow of charge only occurs till both the spheres reach the same potential.