JEE MainPhysicsModerateCapacitance

A fiber sheet with relative permittivity 5 is inserted between
two parallel metal plates 0.25 cm apart. A potential difference of 2500
v is applied between the plates. If dielectric strength of air is
$3 \times 10^{6}v/m$ than

Show answer and solution

V = V_{1} + V_{2}

2500 = 0.0002E_{1} + 0.0023E_{2}

2E_{1} + 23E_{2} = 25 \times 10^{6}

Also

E_{1} = 5E_{2}

Solving

E_{2} = 0757 \times 10^{8}V/m

E_{1} > 3.6 \times 10^{6}

Hence Air will breakdown

Source: Grass-G / P-17 Physics paper 26 Nov Second.docx

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