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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 33 of 37
Marks: +1, -0
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 V m−1m^{-1}. For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 pF?
Solution:  
V = 1 × 10310^3 V , K = 3 , C = 50 × 10−1210^{-12} F
Given E = 10% of dielectric strength
or E = 10100\frac{10}{100} × 4107410^7 V m−1m^{-1} = 106 V m−110^6\,\mathrm{V}\,\mathrm{m}^{-1}
As V = E.d , so d = VE\frac{V}{E} = 13106\frac{1^3}{10^6} = 10−310^{-3} m
and C = kε0Ad\frac{k\varepsilon_0 A}{d}
or A = C⋅dkε0\frac{C\cdot d}{k\varepsilon_0} = 50×10−12×10−33×8.85×10−12\frac{50\times 10^{-12}\times 10^{-3}}{3\times 8.85\times 10^{-12}} or A = 1.9 × 10−3 m210^{-3}\,\mathrm{m}^2 = 19 cm2\mathrm{cm}^2
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