NEET UG 2009 — Physics Electromagnetism

Here, $$\begin{aligned} & \mathrm{V}_{\mathrm{A}}=\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{a}^2}{\mathrm{a}}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{b}^2}{\mathrm{~b}} \\ & +\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi c^2}{c} \\ & =\frac{\sigma}{\varepsilon_0}(a-b+c)=\frac{\sigma}{\varepsilon_0}(2 a) \quad(\because c=a+b) \\ & \end{aligned}$$​VA​=4πε0​1​aσ4πa2​−4πε0​1​ bσ4πb2​+4πε0​1​⋅cσ4πc2​=ε0​σ​(a−b+c)=ε0​σ​(2a)(∵c=a+b)​

$$\begin{aligned} & \mathrm{V}_{\mathrm{B}}=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi \mathrm{a}^2}{\mathrm{a}}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi \mathrm{b}^2}{\mathrm{~b}} \\ &+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi \mathrm{c}^2}{\mathrm{c}} \end{aligned}$$​VB​=4πε0​1​⋅aσ4πa2​−4πε0​1​ bσ4πb2​+4πε0​1​⋅cσ4πc2​​ $$\begin{aligned} & =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-b+c\right)=\frac{\sigma}{\varepsilon_0}(2 a)(\because c=a+b) \\ & \text { and } V_C=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi a^2}{c}-\frac{1}{4 \pi \varepsilon_0} \frac{\sigma 4 \pi b^2}{c} \\ & +\frac{1}{4 \pi \varepsilon_0} \cdot \frac{\sigma 4 \pi c^2}{c} \\ & =\frac{\sigma}{\varepsilon_0}\left(\frac{a^2}{c}-\frac{b^2}{c}+c\right)=\frac{\sigma}{\varepsilon_0}(2 a)(\because c=a+b) \\ & \end{aligned}$$​=ε0​σ​(ca2​−b+c)=ε0​σ​(2a)(∵c=a+b) and VC​=4πε0​1​⋅cσ4πa2​−4πε0​1​cσ4πb2​+4πε0​1​⋅cσ4πc2​=ε0​σ​(ca2​−cb2​+c)=ε0​σ​(2a)(∵c=a+b)​ Hence, $\mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{C}} \neq \mathrm{V}_{\mathrm{B}}$VA​=VC​=VB​