Question

A spherical shell of radius $R$ has a charge $+q$ units. The electric field due to the shell at a point:

& \text{A}\text{. inside is zero and varies as }{{\text{r}}^{\text{-1}}}\text{ outside it} \\\ & \text{B}\text{.inside is constant and varies as }{{\text{r}}^{\text{-2}}}\text{ outside it} \\\ & C.\text{inside is zero and varies as }{{\text{r}}^{\text{-2}}}\text{ outside it} \\\ & \text{D}\text{. inside is constant and varies as }{{\text{r}}^{\text{-1}}}\text{ outside it} \\\ \end{aligned}$$

Answer 1

Electric field is the electric force due to a unit positive charge which is at rest would exert on its surrounding. To calculate the electric field inside the point the shell, we can use gauss law. And to find the electric field outside the shell we can use the formula of electric field.

Formula used:
$E=\dfrac{kq}{r^{2}}$ and $\Phi=\dfrac{q}{\epsilon_{0}}$

Complete step-by-step answer:
We know that the electric force due to a pair of charge is given by the coulombs law. A electric field can be produced by a time-varying electric field or a electrical charge. These can be either attracting or repelling in nature.
An electric field E is defined as the electric force F per unit positive charge q, which is infinitesimally small and at rest, and is given as $E=\dfrac{F}{q}$.
Then$E=\dfrac{kq}{r^{2}}$, where $k=\dfrac{1}{4\pi\epsilon_{0}}$ which is a constant and $r$ is the distance between the unit charges.
Let us consider a spherical shell of radius $R$ with $+q$ charges on the surface area, then the electrical field at any point outside the shell at a distance $r$ from the shell is given by,$E=\dfrac{kq}{r^{2}}$
Hence clearly, $E$ varies with $r^{-2}$.
Also, from Gauss law, we know that the total electric flux through a closed surface is equal to $\dfrac{1}{\epsilon_{0}}$ times the charge enclosed in the surface, and it is given by $\Phi=\dfrac{q}{\epsilon_{0}}$.
We also know that electric flux is the number of electric field lines passing through a unit square area.
Clearly, there is no charge inside the spherical shell; hence we can say there is no electric flux inside the shell. Then clearly, there is no electric field, inside the shell.
Hence clearly, the electric field inside the shell is zero. Outside the shell the electric field varies as $r^{-2}$.
Thus the answer is $C.\text{inside is zero and varies as }{{\text{r}}^{\text{-2}}}\text{ outside it}$

So, the correct answer is “Option C”.

Note: Electric field is in the direction of the force. Usually, the electric field of a point positive charge is radially outwards, whereas the electric field of a point negative charge is radially inwards to the charge. However, the electric field also depends on the symmetry of the charge carrying conductor.