JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Capacitance

A parallel plate capacitor with plates of area

1 m^{2}

each, are at a separation of 0.1 m . If the electric field between the plates is

100 N / C

, the magnitude of charge on each plate is:

(A) $7.85 \times 1 0^{- 10} C$
(B) $6.85 \times 1 0^{- 10} C$
(C) $8.85 \times 1 0^{- 10} C$ ✓ Correct
(D) $9.85 \times 1 0^{- 10} C$

Answer: (C)

Solution

Between the plates the field is

E = \frac{σ}{ε _{0}} = \frac{q}{A ε _{0}}

, so the charge on each plate is

q = E A ε_{0}

. Putting in

E = 100

N/C,

A = 1

^{2}

and

ε_{0} = 8.85 \times 1 0^{- 12}

F/m gives

q = 100 \times 1 \times 8.85 \times 1 0^{- 12} = 8.85 \times 1 0^{- 10}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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