Question 893: Electrostats & Capacitance

Question

A charged conductor produces an electric field of intensity $10^3 \mathrm{~V} / \mathrm{m}$ just outside its surface in vacuum. Then, it produces the electric field of intensity E just outside its surface, when it is placed in a medium of dielectric constant 4. The value of $E$ will be

Accepted Answer

Given that a charged conductor produces an electric field intensity of $10^3 \, \mathrm{V/m}$ just outside its surface in a vacuum, we need to determine the electric field $E$ when the conductor is placed in a medium with a dielectric constant of 4.
The electric field in a medium with a dielectric constant $K$ is given by:
$  E = \frac{E_0}{K}  $
Substituting the given values:
$  E = \frac{10^3}{4} = 250 \, \mathrm{V/m}  $
Thus, the electric field intensity just outside the surface of the conductor when placed in the medium is $250 \, \mathrm{V/m}$.

Answer

$400 \mathrm{~V} / \mathrm{m}$

Answer

$450 \mathrm{~V} / \mathrm{m}$

Answer

$250 \mathrm{~V} / \mathrm{m}$

Answer

$150 \mathrm{~V} / \mathrm{m}$

Electrostats & Capacitance

A charged conductor produces an electric field of intensity 10^3 \mathrm{~V} / \mathrm{m} just outside its surface in vacuum. Then, it produces the electric field of intensity E just outside its surface, when it is placed in a medium of dielectric constant 4. The value of E will be

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