Answer to Question #216988 in Electricity and Magnetism for ArceliaVerana

Question

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

Answer 1

Gives

Solid conducting Sphere

Gauss law

$\phi=\frac{q}{\epsilon_0}$

$E.A=\frac{q}{\epsilon_0}$

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

$E_1=\frac{kQ}{r^2}\\r=a\\E_1=\frac{2kQ}{a^2}$

Part(b)

$\phi=\oint E.dA$

$\phi=E(4\pi r^2)$

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

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

$\phi=\oint E.dA$

$\phi=E(4\pi b^2)$

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

$E_2=\frac{kQa}{r^3}$

$\phi=\oint E.dA$

$\phi=E.(4\pi c^2)$

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

$E_3=\frac{kQr}{c^3}$

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

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

$E_4=\frac{kQ}{c^2}$