A positive point charge q is placed at a dist | vedclass

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(B) Inside a metallic conductor,the net electric field must be zero in electrostatic equilibrium.Let $\vec{E}_{ext}$ be the electric field at the cent

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{q}{9 R^2}$

Vedclass · Answer

(B) Inside a metallic conductor,the net electric field must be zero in electrostatic equilibrium.Let $\vec{E}_{ext}$ be the electric field at the centre of the shell due to the external point charge $q$.Let $\vec{E}_{ind}$ be the electric field at the centre of the shell due to the induced charges on the surface of the shell.Since the net electric field at the centre is zero,we have $\vec{E}_{ext} + \vec{E}_{ind} = 0$,which implies $\vec{E}_{ind} = -\vec{E}_{ext}$.The magnitude of the electric field due to the point charge $q$ at the centre of the shell (which is at a distance $R + 2R = 3R$ from the charge) is given by $E_{ext} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{(3R)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{9R^2}$.Therefore,the magnitude of the electric field due to the induced charge is $|\vec{E}_{ind}| = |\vec{E}_{ext}| = \frac{1}{4 \pi \varepsilon_0} \frac{q}{9R^2}$.

$A$ positive point charge $q$ is placed at a distance $2R$ from the surface of a metallic shell of radius $R$. The electric field at the centre of the shell due to the induced charge has a magnitude of:

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