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Standard XII
Physics
Dielectrics and Polarisation
Question

Two parallel plates of area 100 cm$$^2$$ are given charges of equal magnitudes $$8.9\times10^{-7} C$$ but opposite signs. The electric field within the dielectric material filling the space between the plates is $$1.4 \times 10^6$$ 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 :
$$ κ=\dfrac{q}{ε_0EA}=\dfrac{8..9\times10^{-7}C}{(8.85\times10^{-12}C^2/N.m^2)(1.4\times10^{-6}V/m)(100\times10^{-4}m^2)}=7.2 $$
(b) The charge induced is $$q'=q\left(1-\dfrac{1}{κ}\right)=(8.9\times10^{-7}C)\left(1-\dfrac{1}{7.2}\right)=7.7\times10^{-7}C.$$

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