Question

What is the electric field inside the cavity of any conductor?
A) Infinite
B) $E = \dfrac{{{\varepsilon _0}}}{\sigma }$
C) Zero
D) $E = K\dfrac{\sigma }{{{\varepsilon _0}}}\hat n$

Answer 1

Hint: When the conductor is placed in the electric field the electrons inside the conductor move in the opposite direction of the applied field and create their own electric field. The net charge inside a conductor is zero. Now, apply the gauss’s theorem to get the answer.

Complete step by step answer:
A charge is a massive particle or element which shows electric or magnetic or both effects around it.
The space surrounding the charged particle of a system of charged particles in which we experience the electric effect. It is an electric property associated with each point where a charge is present in any form. It is also defined as an electrical force per unit charge. It is denoted by $E.$
Mathematically, it can be expressed as
$E = \dfrac{F}{Q}$ (where, $F$ is the force and $Q$ is the charge of one particle)
We know that, from Coulomb’s law
$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{qQ}}{{{r^2}}}$
Now, putting this value in formula of electric field, we get
$\Rightarrow E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$
The unit for electric field is $N{C^{ - 1}}$.
When a conductor is placed in the electric field the electrons which are free inside the conductor move opposite to the direction of the applied field. This redistribution creates its own field. So, the charge inside a conductor is zero because they both cancel each other. As the charge inside a conductor is zero. Therefore, if we apply Gauss’s theorem, we get the electric field as zero.

Hence, option (C) is the correct answer.

Note:
Gauss’s Law states that the surface integral of electric field over the closed surface is always equal to the $\dfrac{1}{{{\varepsilon _0}}}$ times of net charge enclosed by the surface.
$\phi = \oint {\vec E.d\vec s = \dfrac{1}{{{\varepsilon _0}}}\left( {{q_{inside}}} \right)} $
Here, $\phi $ is the electric flux and $\phi = ES\cos \theta $.