Two spherical conductors A and B of radii a a | vedclass

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(D) When the inner sphere $A$ (radius $a$) is grounded,its potential $V_A$ becomes $0$.Let the charge on sphere $A$ be $q$ and the charge on sphere $B

Vedclass · Accepted Answer

$4\pi \varepsilon_0 \frac{b^2}{b - a}$

Vedclass · Answer

(D) When the inner sphere $A$ (radius $a$) is grounded,its potential $V_A$ becomes $0$.Let the charge on sphere $A$ be $q$ and the charge on sphere $B$ be $Q$.The potential of sphere $A$ is given by $V_A = \frac{1}{4\pi \varepsilon_0} \frac{q}{a} + \frac{1}{4\pi \varepsilon_0} \frac{Q}{b} = 0$.From this,we get $q = -Q \frac{a}{b}$.The potential of sphere $B$ is $V_B = \frac{1}{4\pi \varepsilon_0} \frac{q}{b} + \frac{1}{4\pi \varepsilon_0} \frac{Q}{b}$.Substituting $q = -Q \frac{a}{b}$ into the equation for $V_B$:$V_B = \frac{1}{4\pi \varepsilon_0} \left( -Q \frac{a}{b^2} + \frac{Q}{b} \right) = \frac{Q}{4\pi \varepsilon_0 b} \left( 1 - \frac{a}{b} \right) = \frac{Q}{4\pi \varepsilon_0 b} \left( \frac{b - a}{b} \right) = \frac{Q(b - a)}{4\pi \varepsilon_0 b^2}$.The capacitance $C$ is defined as $C = \frac{Q}{V_B}$.Therefore,$C = \frac{Q}{\frac{Q(b - a)}{4\pi \varepsilon_0 b^2}} = \frac{4\pi \varepsilon_0 b^2}{b - a}$.

Two spherical conductors $A$ and $B$ of radii $a$ and $b$ $(b > a)$ are placed in air concentrically. $B$ is given a charge $+Q$ and $A$ is grounded. The equivalent capacitance of this system is:

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