Question #216988

A solid conducting sphere of radius a carries a net positive

charge +2Q. A conducting spherical shell of inner radius b and

outer radius c is concentric with the solid sphere and carries a

net charge -Q. Using Gauss's law, find the electric field in the

regions labeled 1, 2, 3, and 4 in Figure and the charge

distribution on the shell when the entire system is in

electrostatic equilibrium


picture : https://ibb.co/m6h9Wpj



1
Expert's answer
2021-07-14T13:41:48-0400

Gives

Solid conducting Sphere




Gauss law

ϕ=qϵ0\phi=\frac{q}{\epsilon_0}

E.A=qϵ0E.A=\frac{q}{\epsilon_0}

E(4πr2)=2Qϵ0E(4\pi r^2)=\frac{2Q}{\epsilon_0}


E1=kQr2r=aE1=2kQa2E_1=\frac{kQ}{r^2}\\r=a\\E_1=\frac{2kQ}{a^2}

Part(b)

ϕ=E.dA\phi=\oint E.dA

ϕ=E(4πr2)\phi=E(4\pi r^2)

4πr2E=Qr3ϵa34\pi r^2 E=\frac{Q r^3}{\epsilon a^3}

E=kQra3E=\frac{kQr}{a^3}

ϕ=E.dA\phi=\oint E.dA

ϕ=E(4πb2)\phi=E(4\pi b^2)

4πa2E=Qa3ϵr34\pi a^2 E=\frac{Q a^3}{\epsilon r^3}

E2=kQar3E_2=\frac{kQa}{r^3}

ϕ=E.dA\phi=\oint E.dA

ϕ=E.(4πc2)\phi=E.(4\pi c^2)

4πc2E=Qc3ϵr34\pi c^2 E=\frac{Q c^3}{\epsilon r^3}

E3=kQrc3E_3=\frac{kQr}{c^3}

E.dA=Qϵ0\oint E.dA=\frac{Q}{\epsilon_0}

E.(4πc2)=Qϵ0E=kQc2E.(4\pi c^2)=\frac{Q}{\epsilon_0}\\E=\frac{kQ}{c^2}

E4=kQc2E_4=\frac{kQ}{c^2}


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