A parallel-plate capacitor has plates of area 0.40 m2 and plate separation of 0.20 mm. The capacitor is connected to a 9.0 V battery. (a) What is the electric field between the plates?(b) What is the capacitance of the capacitor? (c) What is the magnitude of the charge on each plate of the capacitor?

a) E = 4.5*10⁴ V/m

b) C= 17.7 nF

c) Q = 159. 3 nC  

Explanation:

a)

By definition, the electric field is the electrostatic force per unit charge, and since the potential difference between plates is just the work done by the field, divided by the charge, assuming a uniform electric field, if V is the potential difference between plates, and d is the separation between plates, the electric field can be expressed as follows:

       

b)

For a parallel-plate capacitor, applying the definition of capacitance as the quotient between the charge on one of the plates and the potential difference between them, and assuming a uniform surface charge density σ, we get:

       

        From (1), we know that V = E*d, but at the same time, applying Gauss'

        Law at a closed surface half within the plate, half outside it , it can be

        showed than E= σ/ε₀, so finally we get:

       

c)    

From (3) we can solve for Q as follows:

       

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