Question

A hollow spherical shell carries charge density 8 in a region a <r<b. where k is a constant. Find the electric field in the three regions (i) r< a (ii a < r< b,iir >b. Use Gauss's Law For the problem above with the charge distribution Find the potential at the center using infinity as your reference point. V(b)-V(a) =-1,E.dl

Answer 1

Gauss's law states that :    $\int _{closed\, surface}\vec{E}\cdot \vec{dA}=\frac{q_{inside}}{\epsilon _{0}}$

Let the Gaussian surface be a sphere concentric to the given spheres with radius r at which we want to evaluate the field.

For region r<a, the Gaussian surface has no charge inside it so from the Gauss's law the flux of the Electric field through that surface is 0.As the surface is a sphere, Electric field would have been in the radial direction from symmetry, making the total flux through the surface as a positive quantity.But as the flux is zero, the Electric field at the surface has to be 0.

For region r<a ,Electric field at any point is zero.

For region a<r<b ,the Electric field at any point on the Gaussian sphere of radius r is radially outward by symmetry.The Electric charge inside that surface is given by:

Inside a differential hollow sphere at radius x the charge is

$dq=4*\pi*x^2*\rho *dx=4*\pi*x^2*\frac{k}{x^2}*dx=4*\pi*k*dx$

The charge inside sphere of radius r is the sum of all such charges:

$\int _{hollowsphere}dq=\int_{a}^{r}4*\pi*k*dx\, \, ,\, q=4*\pi*k*(r-a)$

Applying Gauss's law we get:

$E*(4*\pi *r^2)=\frac{4*\pi*k*(r-a)}{\epsilon _{0}}\, \, ,\, \, E=\frac{k*(r-a)}{\epsilon _{0}*r^2}$

Electric field in the region a<r<b is given by   $E=\frac{k*(r-a)}{\epsilon _{0}*r^2}$

For region r>b, the Electric field at any point on the Gaussian sphere of radius r is radially outward by symmetry.The Electric charge inside the Gaussian surface is constant and is equal to the entire charge in the sphere.It is given by:

$\int _{hollowsphere}dq=\int_{a}^{b}4*\pi*k*dx\, \, ,\, q=4*\pi*k*(b-a)$

Applying Gauss's law we get:

$E*(4*\pi *r^2)=\frac{4*\pi*k*(b-a)}{\epsilon _{0}}\, \, ,\, \, E=\frac{k*(b-a)}{\epsilon _{0}*r^2}$

Electric field in the region r>b is given by   $E=\frac{k*(b-a)}{\epsilon _{0}*r^2}$

Electric potential with respect to infinity is given by: $V(r)-V(\infty )=-\int_{\infty}^{r}\vec{E}\cdot \vec{dr}$

Thus Electric potential at the center is given by: $V(0)-V(\infty )=-\int_{\infty}^{0}E*dr=\int_{0}^{\infty}E*dr\, \, ,\, \, V(0)-0=\int_{0}^{a}E*dr+\int_{a}^{b}E*dr+\int_{b}^{\infty}E*dr$

Using the Electric field evaluated in the various regions gives:

$V(0)=\int_{0}^{a}0*dr+\int_{a}^{b} \frac{k*(r-a)}{\epsilon _{0}*r^2}*dr+\int_{b}^{\infty}\frac{k*(b-a)}{\epsilon _{0}*r^2}*dr$

$V(0)=\int_{a}^{b} (\frac{k}{\epsilon _{0}*r}-\frac{k*a}{\epsilon _{0}*r^2})*dr+\frac{k*(b-a)}{\epsilon _{0}}*[\frac{-1}{r}]|^{\infty }_{b}$

$V(0)=\frac{k}{\epsilon _{0}}*\ln \frac{b}{a}-(\frac{k*a}{\epsilon _{0}}*\frac{-1}{r})|^{b}_{a}+\frac{k*(b-a)}{\epsilon _{0}*b}$

$V(0)=\frac{k}{\epsilon _{0}}*\ln \frac{b}{a}-\frac{k*(b-a)}{\epsilon _{0}*b}+\frac{k*(b-a)}{\epsilon _{0}*b}$

$\therefore V(0)=\frac{k}{\epsilon _{0}}*\ln \frac{b}{a}$

Potential at the center of the sphere is given by $V(0)=\frac{k}{\epsilon _{0}}*\ln \frac{b}{a}$