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Physics Question on Gauss Law

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

A

B

C

D

Answer

Explanation

Solution

Here, E=0E =0 for 0<r<rA0< r < r _{ A } (as field inside shell is zero) By Gauss's law, the field EE for rA<r<rBr_{A}< r< r_{B} is E.4πr2=Qen ϵ0=QAϵ0E .4 \pi r^{2}=\frac{Q_{\text {en }}}{\epsilon_{0}}=\frac{Q_{A}}{\epsilon_{0}} or E=QA4πϵ0r2E=\frac{Q_{A}}{4 \pi \epsilon_{0} r^{2}} for rA<r<rBr_{A}< r< r_{B} By Gauss's law, the field EE for r>rBr>r_{B} is E.4πr2=Qenϵ0=QAQBϵ0E .4 \pi r^{2}=\frac{Q_{e n}}{\epsilon_{0}}=\frac{Q_{A}-Q_{B}}{\epsilon_{0}} E=QAQB4πϵ0r2E=\frac{Q_{A}-Q_{B}}{4 \pi \epsilon_{0} r^{2}} for r>rBr>r_{B} As QB>QA,E\left| Q _{ B }\right|>\left| Q _{ A }\right|, E is negative for r>rBr > r _{ B }.