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A thin metallic spherical shell of radius $r$ contains a charge $Q$ on its surface. A point charge $q_1$ is placed at the centre of shell and another charge $q_2$ is placed outside the shell at a distance $x$ from the centre. Then, the forces on charges $q_1$ and $q_2$ respectively are
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The correct answer is:
$0, \frac{q_2}{4 \pi \varepsilon_0} \frac{Q+q_1}{x^2}$
Given situation is
As field inside shell is zero, so force on $q_1=q_1 E_{\mathrm{in}}=0$. Also force on $q_2$ is due to both $q_1$ and $Q$.
$F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2 Q}{x^2}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2 q_1}{x^2}$
$\therefore \quad F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2\left(Q+q_1\right)}{x^2}$
As field inside shell is zero, so force on $q_1=q_1 E_{\mathrm{in}}=0$. Also force on $q_2$ is due to both $q_1$ and $Q$.
$F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2 Q}{x^2}+\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2 q_1}{x^2}$
$\therefore \quad F=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q_2\left(Q+q_1\right)}{x^2}$
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