Physics - Electrostatics Question with Solution | TestHub

Let the surface charge density is$\sigma$

$\therefore \text{Q}={\text{q}}_{\text{A}}+{\text{q}}_{\text{B}}$

or$\text{Q}=4\pi {\text{r}}^2\sigma +4\pi {\text{R}}^2\sigma$

$\therefore \sigma =\frac{\text{Q}}{4\pi \left({\text{r}}^2+{\text{R}}^2\right)}$

$\therefore {\text{V}}_{\text{C}}={\text{V}}_{\text{A}}+{\text{V}}_{\text{B}}$

$\therefore {\text{V}}_{\text{C}}=\frac{{\text{q}}_{\text{A}}}{4\pi {\epsilon }_0\text{r}}+\frac{{\text{q}}_{\text{B}}}{4\pi {\epsilon }_0\text{R}}$

$\frac{1}{4\pi {\epsilon }_0}\left[\frac{{\text{q}}_{\text{A}}}{\text{r}}+\frac{{\text{q}}_{\text{B}}}{\text{R}}\right]=\text{k}\left(\frac{4\pi {\text{r}}^2\sigma }{\text{r}}+\frac{4\pi {\text{R}}^2\sigma }{\text{R}}\right)$

$\text{k}=\left(4\pi \text{r}\sigma +4\pi \text{R}\sigma \right)$

$=4\pi \text{k}\left[\frac{\text{Qr}}{4\pi \left({\text{R}}^2+{\text{r}}^2\right)}+\frac{\text{QR}}{4\pi \left({\text{r}}^2+{\text{R}}^2\right)}\right]$

$=\frac{\text{kQ}\left(\text{R}+\text{r}\right)}{ \left({\text{R}}^2+{\text{r}}^2\right)}\left(\text{Here}\text{, }\text{k}=\frac{1}{4\pi {\epsilon }_0}\right)$