Question

A charge Q is distributed uniformly over a metallic sphere of radius R. Obtain the expressions for the electric field (E) and electric potential (V) at a point 0 < x < R.
Show on a plot the variation of E and V with x for 0 < x < 2R.

Answer 1

Suppose, we have to calculate the electric field at the point P at a distance x (x < R) from its centre. Draw the Gaussian surface through point P to enclose the charged spherical shell. The Gaussian surface is a spherical shell of radius x and centre O.

Let `vecE` be the electric field at point P. Then, the electric flux through area element `vec(ds)`

`dø = vecE.vec(ds) `

Since `vec(dS)` is also along the normal to the surface, \[d\phi = Eds\]

∴ Total electric flux through the Gaussian surface,

Now,

\[\oint dS = 4 \pi x^2 \]

\[ \therefore \phi = E \times 4 \pi x^2 . . . . . (i)\]

and according to Gauss's theorem,  

\[\phi = \frac{Q}{\epsilon_o}\]

where Q is the charge enclosed by the gaussian surface.
So, 

\[E = \frac{Q}{4\pi \epsilon_0}\frac{1}{x^2}\]

As charge inside a metallic sphere is 0

\[\Rightarrow\]E = 0

i.e. E = 0   (x < R)
We know that, 

\[E = - \frac{dV}{dx}\]

\[\Rightarrow dV = - Edx\]

\[ \Rightarrow V = - \int_0^R E d x\]

\[ \Rightarrow V = - \int_0^R \frac{Q}{4\pi \epsilon_0 x^2} d x\]

\[ \Rightarrow V = \frac{Q}{4\pi \epsilon_0 R}\]

At every point within the sphere potential is same as that on surface. 

Plot of variation of E with x for 0 < x < 2R

Plot of variation of V with x for 0 < x < 2R