Q વિદ્યુતભાર ધરાવતા એક ધન વાહક ગોળોએ અવિદ્યુત | vedclass

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Question

(a) In case of a charged conducting sphere
${V_{{\rm{inside}}}} = {V_{{\rm{centre }}}} = {V_{{\rm{surface}}}} = \frac{1}{{4\pi {\varepsilon _o}}}.\fr

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(a) In case of a charged conducting sphere
${V_{{m{inside}}}} = {V_{{m{centre }}}} = {V_{{m{surface}}}} = \frac{1}{{4\pi {\varepsilon _o}}}.\frac{q}{R}$, ${V_{{m{outside}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{q}{r}$
If $a$ and $b$ are the radii of sphere and spherical shell respectively, then potential at their surface will be
${V_{{m{sphere }}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{a}$ and ${V_{{m{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{Q}{b}$
$	herefore $$V = {V_{{m{sphere }}}} - {V_{{m{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}.\left[ {\frac{Q}{a} - \frac{Q}{b}} ight]$
Now when the shell is given charge $(-3Q)$, then the potential will be
$V{'_{{m{sphere}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{a} + \frac{{( - 3Q)}}{b}} ight],$$V{'_{{m{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{b} + \frac{{( - 3Q)}}{b}} ight]$
$	herefore $$V{'_{{m{sphere }}}} - V{'_{{m{shell}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\left[ {\frac{Q}{a} - \frac{Q}{b}} ight] = V$

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