Question:

Two concentric conducting thin spherical shells $A$ and $B$ having radii $r _{ A }$ and $r _{ B }\left( r _{ B }> r _{ A }\right)$ are charged to $Q _{ A }$ and $- Q _{ B }\left(\left| Q _{ B }\right|>\left| Q _{ A }\right|\right)$. The electrical field along a line passing through the centre is

Updated On: Jul 14, 2022
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The Correct Option is A

Solution and Explanation

Here, $E =0$ for $0< r < r _{ A } $ (as field inside shell is zero) By Gauss's law, the field $E$ for $r_{A}< r< r_{B}$ is $E .4 \pi r^{2}=\frac{Q_{\text {en }}}{\epsilon_{0}}=\frac{Q_{A}}{\epsilon_{0}}$ or $E=\frac{Q_{A}}{4 \pi \epsilon_{0} r^{2}}$ for $r_{A}< r< r_{B}$ By Gauss's law, the field $E$ for $r>r_{B}$ is $E .4 \pi r^{2}=\frac{Q_{e n}}{\epsilon_{0}}=\frac{Q_{A}-Q_{B}}{\epsilon_{0}}$ $E=\frac{Q_{A}-Q_{B}}{4 \pi \epsilon_{0} r^{2}}$ for $r>r_{B}$ As $\left| Q _{ B }\right|>\left| Q _{ A }\right|, E$ is negative for $r > r _{ B }$.
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Concepts Used:

Gauss Law

Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.

Gauss Law:

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.

Gauss Law Formula:

As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;

Q = ϕ ϵ0

The Gauss law formula is expressed by;

ϕ = Q/ϵ0

Where,

Q = total charge within the given surface,

ε0 = the electric constant.