Question

An electric field of 9.50×105 V/m is desired between two parallel plates, each of area 45.0 cm2 and separated by 2.45 mm of air. What charge must be on each plate?

Answer 1

Solution)

Given,

Electric Field, E=9.5*10^5 V/m

Area, A=45*10^-4 m^2

Distance, d=2.45*10^-3 m

We know, Capacitance, C= AEo/d

Substitute values,

C=(45*10^-4)*(8.85*10^-12)/(2.45*10^-3)=1.6*10^-11 F

We know, Electric Potential, V= Ed=(9.5*10^5)*(2.45*10^-3)

=2327.5 V

Now, Charge, Q=C*V

Substitute values,

Q=(1.6*10^-11)*(2327.5)= 3.72*10^-8 C (Ans)

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