Physics - Electrostatics Question with Solution | TestHub

Here, $$\begin{array}{rl}& {\mathrm{V}}_{\mathrm{A}}=\frac{1}{4\pi {\epsilon }_0}\frac{\sigma 4\pi {\mathrm{a}}^2}{\mathrm{a}}-\frac{1}{4\pi {\epsilon }_0}\frac{\sigma 4\pi {\mathrm{b}}^2}{\mathrm{b}}\\ & +\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\sigma 4\pi c^2}{c}\\ & =\frac{\sigma }{{\epsilon }_0}(a-b+c)=\frac{\sigma }{{\epsilon }_0}(2a)(\because c=a+b)\\ & \end{array}$$
$$\begin{array}{rl}& {\mathrm{V}}_{\mathrm{B}}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\sigma 4\pi {\mathrm{a}}^2}{\mathrm{a}}-\frac{1}{4\pi {\epsilon }_0}\frac{\sigma 4\pi {\mathrm{b}}^2}{\mathrm{b}}\\ & +\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\sigma 4\pi {\mathrm{c}}^2}{\mathrm{c}}\end{array}$$ $$\begin{array}{rl}& =\frac{\sigma }{{\epsilon }_0}\left(\frac{a^2}{c}-b+c\right)=\frac{\sigma }{{\epsilon }_0}(2a)(\because c=a+b)\\ & \text{ and }{V}_C=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\sigma 4\pi a^2}{c}-\frac{1}{4\pi {\epsilon }_0}\frac{\sigma 4\pi b^2}{c}\\ & +\frac{1}{4\pi {\epsilon }_0}\cdot \frac{\sigma 4\pi c^2}{c}\\ & =\frac{\sigma }{{\epsilon }_0}\left(\frac{a^2}{c}-\frac{b^2}{c}+c\right)=\frac{\sigma }{{\epsilon }_0}(2a)(\because c=a+b)\\ & \end{array}$$ Hence, ${\mathrm{V}}_{\mathrm{A}}={\mathrm{V}}_{\mathrm{C}}\ne {\mathrm{V}}_{\mathrm{B}}$