Practice portal › Electric Potential and Capacitance › Capacitance and Dielectrics

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3×10⁴ V m⁻¹, the charge density of the positive plate will be close to

Asked in JEE Main 2014 · Polarisation and bound charge

Answer: (2) 6×10⁻⁷ C m⁻²

Step-by-step solution

Inside a dielectric the free charge density on the plate is fixed by E=σ/(Kε₀), so σ=Kε₀E.

Kε₀=2.2×8.85×10⁻¹²=1.947×10⁻¹¹.

σ=1.947×10⁻¹¹×3×10⁴=5.84×10⁻⁷ C m⁻².

This is close to 6×10⁻⁷ C m⁻². (The plate separation of 5 mm is not needed.)

Why the other options are wrong

More Capacitance and Dielectrics questionsAll Capacitance and Dielectrics questions →

Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.

Practise this with a timer