Question

An electric charge produces an electric intensity of $500\,N{C^{ - 1}}$ at a point in air. If the air is replaced by a medium of dielectric constant $2.5$, then the intensity of the electric field due to the same charge at the same point will be:
(A) $100\,N{C^{ - 1}}$
(B) $150\,N{C^{ - 1}}$
(C) $200\,N{C^{ - 1}}$
(D) $300\,N{C^{ - 1}}$

Answer 1

Hint: The intensity of the electric field due to the same charge at the same point can be determined by using the formula of the Gauss law. Gauss law gives the relation between the electric field intensity and the total charge in the system and the electric constant.

Useful formula
The Gauss’s law formula is given by,
${\phi _E} = \dfrac{Q}{{{\varepsilon _0}}}$
Where, ${\phi _E}$ is the electric flux or electric field intensity, $Q$ is the total electric charge and ${\varepsilon _0}$ is the electric constant of the medium.

Complete step by step solution
Given that,
The electric charge of the system is, $Q = 500\,N{C^{ - 1}}$
The dielectric constant of the medium is, ${\varepsilon _0} = 2.5$
Now, the intensity of the electric field due to the same charge at the same point will be given by,
Gauss's law shows that the relation between the electric field intensity and the total charge in the system and the electric constant. Then,
${\phi _E} = \dfrac{Q}{{{\varepsilon _0}}}\,............................\left( 1 \right)$
By substituting the electric charge of the system and the dielectric constant of the medium in the above equation (1), then the above equation (1) is written as,
${\phi _E} = \dfrac{{500}}{{2.5}}$
By dividing the terms in the above equation, then the above equation is written as,
${\phi _E} = 200\,N{C^{ - 1}}$
Thus, the above equation shows the intensity of the electric field due to the same charge at the same point, when the charge is placed in the medium of dielectric instead of the air medium.

Hence, the option (C) is the correct answer.

Note: The electric field intensity produced is directly proportional to the charge of the system and the electric field intensity is inversely proportional to the electric constant of the medium. If the charge increases, then the intensity also increases. If the electric constant increases, then the intensity decreases.