A thin metallic spherical shell of radius r c | vedclass

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(D) $1$. The electric field inside a conducting spherical shell is zero because the charges on the shell redistribute themselves to cancel any interna

Vedclass · Accepted Answer

$0, \frac{q_2}{4 \pi \varepsilon_0} \frac{Q+q_1}{x^2}$

Vedclass · Answer

(D) $1$. The electric field inside a conducting spherical shell is zero because the charges on the shell redistribute themselves to cancel any internal field. Therefore, the force on the charge $q_1$ placed at the centre is $F_1 = q_1 	imes E_{in} = q_1 	imes 0 = 0$.$2$. For the charge $q_2$ placed outside the shell at a distance $x$ from the centre, the shell acts as a point charge $Q$ located at its centre (by Gauss's Law). Additionally, the charge $q_1$ at the centre also exerts a force on $q_2$.$3$. The total electric field at the position of $q_2$ is the sum of the fields produced by $Q$ and $q_1$, which is $E_{total} = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{x^2} + \frac{1}{4 \pi \varepsilon_0} \frac{q_1}{x^2} = \frac{1}{4 \pi \varepsilon_0} \frac{Q+q_1}{x^2}$.$4$. The force on $q_2$ is $F_2 = q_2 	imes E_{total} = \frac{q_2}{4 \pi \varepsilon_0} \frac{Q+q_1}{x^2}$.

$A$ 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 the 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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