JEE Main · 2014 · 06 Apr 2014 · Offline · Capacitance

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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 \times 1 0^{4} V / m

, the charge density of the positive plate will be close to :

(A) $3 \times 1 0^{4} C / m^{2}$
(B) $6 \times 1 0^{4} C / m^{2}$
(C) $6 \times 1 0^{- 7} C / m^{2}$ ✓ Correct
(D) $3 \times 1 0^{- 7} C / m^{2}$

Answer: (C)

Solution

Inside the dielectric the field and the free charge density on the plate are tied by

E = \frac{σ}{K ε _{0}}

, so

σ = K ε_{0} E

. Putting in the numbers,

σ = (2.2) (8.85 \times 1 0^{- 12}) (3 \times 1 0^{4}) \approx 5.8 \times 1 0^{- 7}

, which rounds to about

6 \times 1 0^{- 7}

C/m

^{2}

JEE Main · 2014 · 06 Apr 2014 · Offline · Single correct option

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