A parallel-plate capacitor with plate area A = 5.5 m2 and plate separation d = 3

Question

A parallel-plate capacitor with plate area A = 5.5 m2 and plate separation d = 3.0 mm is connected to a 35-V battery (refer to (Figure 1) ). Determine the energy stored in the capacitor. With the capacitor still connected to the battery, a slab of plastic with dielectric strength K = 3.2 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric (refer to (Figure 2) ). What are the new value of electric field in the capacitor? What are the new value of the capacitance?What are the new value the energy stored in the capacitor?

Explanation / Answer

Capacitance C = eoA/d

= 8.85*10^-12 *5.5/3mm

= 16.225 nF

Charge Q = CV

Q = 16.225 nF * 35V

Q = 5.679*10^-7 C or 0.568 uC

1. Energy stored U = 0.5 cv^2

U = 0.5 * 16.225 *10^-9 *35*35

U = 9.938 *10^-6 Joules

2. EFnew= Eo

Since the battery is still connected, V=35V

E=dV/dN=(35/3*10-3m)= 1.166*10^4 V/m

3. Cnew = Kco

Cnew = 3.2* 16.225 nF

Cnew = 51.92 nF

4. charge new = CV

= 51.92 * 35

= 1.817 *10^-6 C or 1.817 uC

5. PE new = 0.5 cv^2 = 0.5 * 51.92nF * 35*35

PEnew = 3.18*10-5 J

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