Two metal spheres,one of radius R and the oth | vedclass

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(D) Initial charges on the spheres are $Q_1 = \sigma (4 \pi R^2)$ and $Q_2 = \sigma (4 \pi (2R)^2) = 16 \pi R^2 \sigma$.Total charge $Q_{total} = Q_1

Vedclass · Accepted Answer

$\sigma_{1}=\frac{5}{3} \sigma, \sigma_{2}=\frac{5}{6} \sigma$

Vedclass · Answer

(D) Initial charges on the spheres are $Q_1 = \sigma (4 \pi R^2)$ and $Q_2 = \sigma (4 \pi (2R)^2) = 16 \pi R^2 \sigma$.Total charge $Q_{total} = Q_1 + Q_2 = 4 \pi R^2 \sigma + 16 \pi R^2 \sigma = 20 \pi R^2 \sigma$.When the spheres are brought into contact,the total charge is redistributed such that they reach the same potential $V = \frac{k Q_1'}{R} = \frac{k Q_2'}{2R}$.This implies $Q_2' = 2 Q_1'$.Since $Q_1' + Q_2' = 20 \pi R^2 \sigma$,we have $Q_1' + 2 Q_1' = 20 \pi R^2 \sigma$,which gives $3 Q_1' = 20 \pi R^2 \sigma$,so $Q_1' = \frac{20}{3} \pi R^2 \sigma$.Then $Q_2' = 2 	imes \frac{20}{3} \pi R^2 \sigma = \frac{40}{3} \pi R^2 \sigma$.The new surface charge densities are $\sigma_1' = \frac{Q_1'}{4 \pi R^2} = \frac{20/3 \pi R^2 \sigma}{4 \pi R^2} = \frac{5}{3} \sigma$.And $\sigma_2' = \frac{Q_2'}{4 \pi (2R)^2} = \frac{40/3 \pi R^2 \sigma}{16 \pi R^2} = \frac{40}{48} \sigma = \frac{5}{6} \sigma$.

Two metal spheres,one of radius $R$ and the other of radius $2R$,respectively,have the same surface charge density $\sigma$. They are brought in contact and separated. What will be the new surface charge densities on them?

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