Question bank/JEE Main/Physics
JEE MainPhysicsModerateCapacitance
A fiber sheet with relative permittivity 5 is inserted between
two parallel metal plates 0.25 cm apart. Apotential difference of 2500 v
is applied between the plates. If dielectric strength of air is 3 ×
10^6 v/m then.
Show answer and solution
As V = V_1 + V_2
2500 = 0.0002 E_1 + 0.0023 E_2
2E_1 + 23 E_2 = 25 x 10^6
Also; E_1 = 5E_2
E_1 = 3.788 x 10^6 V/m
As E_1> 3.6 x 10^6 Hence Air will breakdown
2500 = 0.0002 E_1 + 0.0023 E_2
2E_1 + 23 E_2 = 25 x 10^6
Also; E_1 = 5E_2
E_1 = 3.788 x 10^6 V/m
As E_1> 3.6 x 10^6 Hence Air will breakdown
Source: Grass-G / P-12 Physics paper 2 Nov.docx
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