Two parallel plates of area 100 cm $^{2}$ are given charges of equal magnitudes $8.9 \times 1 0^{- 7} C$ but opposite signs. The electric field within the dielectric material filling the space between the plates is $1.4 \times 1 0^{6}$ 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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Answer 1

(a) We apply Gauss’s law with dielectric and solve for k :
$κ = \frac{q}{ε _{0} E A} = \frac{8 . . 9 \times 1 0 ^{- 7} C}{( 8 . 8 5 \times 1 0 ^{- 12} C ^{2} / N . m ^{2} ) ( 1 . 4 \times 1 0 ^{- 6} V / m ) ( 1 0 0 \times 1 0 ^{- 4} m ^{2} )} = 7.2$

(b) The charge induced is $q^{'} = q (1 - \frac{1}{κ}) = (8.9 \times 1 0^{- 7} C) (1 - \frac{1}{7 . 2}) = 7.7 \times 1 0^{- 7} C .$

Answer 2

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

κ = \frac{q}{ε _{0} E A} = \frac{8 . . 9 \times 1 0 ^{- 7} C}{( 8 . 8 5 \times 1 0 ^{- 12} C ^{2} / N . m ^{2} ) ( 1 . 4 \times 1 0 ^{- 6} V / m ) ( 1 0 0 \times 1 0 ^{- 4} m ^{2} )} = 7.2

(b) The charge induced is

q^{'} = q (1 - \frac{1}{κ}) = (8.9 \times 1 0^{- 7} C) (1 - \frac{1}{7 . 2}) = 7.7 \times 1 0^{- 7} C .

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