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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) \( 1.77 \times 10^{-6} C / m ^{2} \) |
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Answer» 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) |
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