तीन समकेन्द्री गोलों की त्रिज्याएं a , b और c | vedclass

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Question

$V_{A}=\frac{1}{4 \pi \varepsilon_{0}}\left\{\frac{q_{A}}{a}+\frac{q_{B}}{b}+\frac{q_{C}}{c}\right\}$

$=\frac{4 \pi}{4 \pi \varepsilon_{0}}\left\{\

Vedclass · Accepted Answer

$V_C=V_A 
e VB$ होगा

Vedclass · Answer

$V_{A}=\frac{1}{4 \pi \varepsilon_{0}}\left\{\frac{q_{A}}{a}+\frac{q_{B}}{b}+\frac{q_{C}}{c}ight\}$

$=\frac{4 \pi}{4 \pi \varepsilon_{0}}\left\{\frac{a^{2} \sigma}{a}-\frac{b^{2} \sigma}{b}+\frac{c^{2} \sigma}{c}ight\}$

${V_{A}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{a}-\frac{b^{2} \sigma}{b}+\frac{c^{2} \sigma}{c}ight\}} $

${V_{B}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{b}-\frac{b^{2} \sigma}{b}+\frac{c^{2} \sigma}{c}ight\}}$

$V_{C}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{c}-\frac{b^{2} \sigma}{c}+\frac{c^{2} \sigma}{c}ight\}$

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}ight\}} $

${V_{B}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{2 a}-\frac{4 a^{2} \sigma}{2 a}+\frac{c^{2} \sigma}{c}ight\}}$

$V_{C}=\frac{1}{\varepsilon_{0}}\left\{\frac{a^{2} \sigma}{3 a}-\frac{4 a^{2} \sigma}{3 a}+\frac{c^{2} \sigma}{c}ight\}$

It can seen by taking out common factors that

$V_{A}=V_{C}>V_{B} \quad 	ext { i.e., } \quad V_{A}=V_{C} 
eq V_{B}$

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