Question

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?

Answer 1

Answer:

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:

       $E = \frac{V}{d} = \frac{9.0V}{2*10-4m} =4.5 * 10e4 V/m (1)$

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:

       $Q = \sigma* A (2)$

        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:

       $C = \frac{Q}{V} =\frac{\sigma*A}{E*d} = \frac{\sigma*A}{\frac{\sigma}{\epsilon_{o} } d} =\frac{\epsilon_{0}*A}{d} = \frac{8.85e-12F/m*0.4m2}{2e-4m} = 17.7 nF (3)$

c)    

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

       $Q = C* V = 17.7 nF * 9.0 V = 159.3 nC (4)$