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Two isolated metallic solid spheres of radii R and 2 R are charged such that both of these have same density\sigma. The spheres are located far away from each other, and connected by a thin conducting wire. The new charge density on the bigger sphere -

Option: 1

\frac{5}{3} \frac{R \sigma}{\epsilon_0}


Option: 2

\frac{4}{g} \frac{R \sigma}{\epsilon_0}


Option: 3

\frac{5}{9} \frac{R \sigma}{t_0}


Option: 4

\frac{3}{5} \frac{R \sigma}{G_0}


Answers (1)

best_answer

After connection, the common potential.
\begin{aligned} V & =\frac{Q_1+Q_2}{C_1+C_2} \\ & =\frac{\left[4 \pi R^2+4 \pi(2 R)^2\right] \sigma}{4 \pi t_0 R+4 \pi \epsilon_0(2 R)} \\ & =\frac{\left[4 \pi R^2+16 \pi R^2\right] \sigma}{12 \pi \epsilon_0 R}=\frac{20 \pi R^2 \sigma}{12 \pi t_0 R} \\ & =\frac{5}{3} \frac{R \sigma}{\epsilon_0} \end{aligned} 

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chirag

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