Question

Two parallel plates of area 100 cm are given charges of equal magnitudes but opposite signs. The electric field within the dielectric material filling the space between the plates is V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface.

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Solution
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(a) We apply Gauss’s law with dielectric and solve for k :


(b) The charge induced is

Solve any question of Electrostatic Potential and Capacitance with:-
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Assertion

The electric susceptibility of a dielectric material is a measure of how easily it polarizes in response to an electric field.

Reason

The electric susceptibility determines the electric permittivity of the material and thus influences many other phenomena in that medium, from the capacitance of capacitors to the speed of light.

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