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Electric Charges and Fields — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectric Charges and Fields5 Marks
Question
  1. Using Gauss Theorem show mathematically that for any point outside the shell, the field due to a uniformly charged spherical shell is same as the entire charge on the shell, is concentrated at the centre.
  2. Why do you expect the electric field inside the shell to be zero according to this theorem?
OR
A thin conducting spherical shell of radius R has charge Q spread uniformly over its surface. Using Gauss’s theorem, derive an expression for the electric field at a point outside the shell.
Draw a graph of electric field E(r) with distance r from the centre of the shell for $0\leq\text{r}\le\infty.$
OR
Find the electric field intensity due to a uniformly charged spherical shell at a point (i) outside the shell and (ii) inside the shell. Plot the graph of electric field with distance from the centre of the shell.
OR
Using Gauss’s law obtain the expression for the electric field due to a uniformly charged thin spherical shell of radius R at a point outside the shell. Draw a graph showing the variation of electric field with r, for r > R and r < R.
✓

Answer

Image
  1. Electric field intensity at a point outside a uniformly charged thin spherical shell: Consider a uniformly charged thin spherical shell of radius R carrying charge Q. To find the electric field outside the shell, we consider a spherical Gaussian surface of radius r (>R), concentric with given shell. If $\vec{\text{E}}$ is electric field outside the shell, then by symmetry electric field strength has same magnitude $E_0$ on the Gaussian surface and is directed radially outward. Also the directions of normal at each point is radially outward, so angle between $\vec{\text{E}}_\text{i}$ and $\vec{\text{dS}}$ is zero at each point. Hence, electric flux through Gaussian surface.
$\oint\text{S}=\vec{\text{E}}_0.\vec{\text{dS}}$
$\oint=\text{E}_0\text{dS}\cos0=\text{E}_0.4\pi\text{r}^2$
Now, Gaussian surface is outside the given charged shell, so charge enclosed by Gaussian surface is Q.
Hence, by Gauss’s theorem,
$\oint\text{S}=\vec{\text{E}}_0.\vec{\text{dS}}=\frac{1}{\in_0}\times\text{charged enclosed}$
$\Rightarrow\ \text{E}_0.4\pi\text{r}^2=\frac{1}{\in_0}\times\text{Q}$
$\Rightarrow\ \text{E}_0=\frac{1}{4\pi\in_0}\frac{\text{Q}}{\text{r}^2}$
Thus, electric field outside a charged thin spherical shell is the same as if the whole charge Q is concentrated at the centre.
If $\sigma$ is the surface charge density of the spherical shell, then
$\text{Q}=4\pi\text{R}^2\sigma$ coulomb
$\therefore\ \text{E}_0=\frac{1}{4\pi\in_0}\frac{4\pi\text{R}^2\sigma}{\text{r}^2}=\frac{\text{R}^2\sigma}{\in_0\text{r}^2}$
Image
  1. Electric field inside the shell (hollow charged conducting sphere): The charge resides on the surface of a conductor. Thus a hollow charged conductor is equivalent to a charged spherical shell. To find the electric field inside the shell, we consider a spherical Gaussian surface of radius r (< R) concentric with the given shell. If $\vec{\text{E}}$ is the electric field inside the shell, then by symmetry electric field strength has the same magnitude $E_i$ on the Gaussian surface and is directed radially outward. Also the directions of normal at each point is radially outward, so angle between $\vec{\text{E}}_\text{i}$ and $\vec{\text{dS}}$ is zero at each point.
Hence, electric flux through Gaussian surface,
$=\int\limits_\text{S}\vec{\text{E}}_{\text{i}}.\vec{\text{dS}}=\int\text{E}_\text{i}.\text{dS}\cos0=\text{E}_\text{i}.4\pi\text{r}^2$
Now, Gaussian surface is inside the given charged shell, so charge enclosed by Gaussian surface is zero.
Hence, by Gauss’s theorem,
$\int\limits_\text{S}\vec{\text{E}}_\text{i}.\vec{\text{dS}}=\frac{1}{\in_0}\times\text{charged enclosed}$
$\Rightarrow\ \text{E}_\text{i}4\pi\text{r}^2=\frac{1}{\in_0}\times0\Rightarrow\ \text{E}_\text{i}=0$
Image
Thus, electric field at each point inside a charged thin spherical shell is zero. The graph is shown in fig.

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