55. Equal and opposite charge is given to two parallel plates. Without any space between them, the electric field is \( 3 \times 10^{5} V / m \) and with space filled with dielectric, electric field intensity is \( 1.0 \times 10^{5} V / m \). The induced charge density on the surface of the dielectric is: (a) \( 3.44 \times 10^{-6} C / m ^{2} \) (b) \( 8.8 \times 10^{-6} C / m ^{2} \) (c) \( 6.93 \times 10^{-6} C / m ^{2} \) (d) \( 1.77 \times 10^{-6} C / m ^{2} \)
55. Equal and opposite charge is given to two parallel plates. Without any space between them, the electric field is \( 3 \times 10^{5} V / m \) and with space filled with dielectric, electric field intensity is \( 1.0 \times 10^{5} V / m \). The induced charge density on the surface of the dielectric is: (a) \( 3.44 \times 10^{-6} C / m ^{2} \) (b) \( 8.8 \times 10^{-6} C / m ^{2} \) (c) \( 6.93 \times 10^{-6} C / m ^{2} \) (d) \( 1.77 \times 10^{-6} C / m ^{2} \)
1 Answers
Given
E = 3 x 105 v/m
E0 = 1.0 x 105 v/m
Dielectric constant
K = E/E0
K = \(\frac{3\times10^5}{1\times10^5}\)
K = 3
\(\because\) E = \(\frac{\sigma}{\varepsilon_0}\)
σ = E \(\varepsilon_0\)
σ = 3 x 105 x 8.85 x 10-12
σ = 26.55 x 10-7
induced charged
σ induced = σ\(\frac{(k-1)}{(k)}\)
σ induced = 26.55 x 10-7 x 2/3
σ induced = 1.77 x 10-6 c/m2
option(d)