બે સમકેન્દ્રીય ગોળીય કવચની ત્રિજયા r અને R (R | vedclass

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Question

$Q = {q_1} + {q_2}$&hellip;.$(i)$${\sigma _1} = {\sigma _2}$$\frac{{{q_1}}}{{4\pi {r^2}}} = \frac{{{q_2}}}{{4\pi {R^2}}}$

$==>$ $\frac{{{q_1}}}{

Vedclass · Accepted Answer

$\frac{{Q(R + r)}}{{4\pi {\varepsilon _0}({R^2} + {r^2})}}$

Vedclass · Answer

$Q = {q_1} + {q_2}$&hellip;.$(i)$${\sigma _1} = {\sigma _2}$$\frac{{{q_1}}}{{4\pi {r^2}}} = \frac{{{q_2}}}{{4\pi {R^2}}}$

$==>$ $\frac{{{q_1}}}{{{q_2}}} = \frac{{{r^2}}}{{{R^2}}}$&hellip;. $(ii)$

${q_1} = \frac{{Q{r^2}}}{{({R^2} + {r^2})}}$ ,${q_2} = \frac{{Q{R^2}}}{{({R^2} + {r^2})}}$

$V = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{{{q_1}}}{r} + \frac{{{q_2}}}{R}} \right]$

$ = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{{Qr}}{{({R^2} + {r^2})}} + \frac{{QR}}{{({R^2} + {r^2})}}} \right] = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{{Q(R + r)}}{{({R^2} + {r^2})}}$

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