$=\frac{4 \pi}{4 \pi \varepsilon_{0}}\left\{\frac{a^{2} \sigma}{a}-\frac{b^{2} \sigma}{b}+\frac{c^{2} \sigma}{c}\right\}$
${V_{A}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{a}-\frac{b^{2} \sigma}{b}+\frac{c^{2} \sigma}{c}\right\}} $
${V_{B}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{b}-\frac{b^{2} \sigma}{b}+\frac{c^{2} \sigma}{c}\right\}}$
$V_{C}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{c}-\frac{b^{2} \sigma}{c}+\frac{c^{2} \sigma}{c}\right\}$
Given $c=a+b$ If $a=a, b=2 a$ and $c=3 a$ for example, as $c>b>a$
${V_{A}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{a}-\frac{4 a^{2} \sigma}{2 a}+\frac{c^{2} \sigma}{c}\right\}} $
${V_{B}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{2 a}-\frac{4 a^{2} \sigma}{2 a}+\frac{c^{2} \sigma}{c}\right\}}$
$V_{C}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{3 a}-\frac{4 a^{2} \sigma}{3 a}+\frac{c^{2} \sigma}{c}\right\}$
It can seen by taking out common factors that
$V_{A}=V_{C}>V_{B} \quad \text { i.e., } \quad V_{A}=V_{C} \neq V_{B}$
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