A change Q is distributed over two concentric conducting thin spherical shells radii r and R (R>r). If the surface charge densities on the two shells are equal, the electric potential at the common centre is ;       
Option: 1 \frac{1}{4\pi \epsilon _0}\frac{\left ( 2R+r \right )}{\left ( R^2+r^2 \right )}Q
 
Option: 2 \frac{1}{4\pi \epsilon _0}\frac{\left (R+r \right )}{2\left ( R^2+r^2 \right )}Q  
Option: 3 \frac{1}{4\pi \epsilon _0}\frac{\left (R+r \right )}{\left ( R^2+r^2 \right )}Q
 
Option: 4 \frac{1}{4\pi \epsilon _0}\frac{\left (R+2r \right )Q}{2\left ( R^2+r^2 \right )}

Answers (1)

\begin{array}{l} \mathrm{Q}_{1}=\sigma 4 \pi \mathrm{r}^{2} \\ \mathrm{Q}_{2}=\sigma 4 \pi \mathrm{R}^{2} \end{array}

Where {Q}_{1}= charge on inner shell and  {Q}_{2}= charge on the outer shell  

and Q={Q}_{1}+{Q}_{2}

So

\begin{array}{l} Q=\sigma 4 \pi\left(r^{2}+R^{2}\right) \\ \sigma=\frac{Q}{4 \pi\left(r^{2}+R^{2}\right)} \end{array}

\begin{array}{l} \mathrm{V}_{\mathrm{C}}=\frac{\mathrm{KQ}_{1}}{\mathrm{r}}+\frac{\mathrm{KQ}_{2}}{\mathrm{R}} =\frac{\mathrm{K} \sigma 4 \pi \mathrm{r}^{2}}{\mathrm{r}}+\frac{\mathrm{K} \sigma 4 \pi \mathrm{R}^{2}}{\mathrm{R}} =\mathrm{K} \sigma 4 \pi(\mathrm{r}+\mathrm{R}) \\ \\ V_C=\frac{\mathrm{KQ} 4 \pi(\mathrm{r}+\mathrm{R})}{4 \pi\left(\mathrm{r}^{2}+\mathrm{R}^{2}\right)} =\frac{\mathrm{KQ}(\mathrm{r}+\mathrm{R})}{\left(\mathrm{r}^{2}+\mathrm{R}^{2}\right)} =\frac{\mathrm{Q}(\mathrm{r}+\mathrm{R})}{4\pi\epsilon _0\left(\mathrm{r}^{2}+\mathrm{R}^{2}\right)}\end{array}

 

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