A parallel-plate capacitor with plate area A = 2.5 m 2 and plate separation d =

Question

A parallel-plate capacitor with plate area A = 2.5m2and plate separation d = 3.0 mm is connected to a 35-V battery (refer to (Figure 1) ).

A parallel-plate capacitor with plate area A = 2.5m2 and plate separation d = 3.0 mm is connected to a 35-V battery (refer to (Figure 1) ). Determine the electric field in the capacitor. E0 = Determine the capacitance. C0 = Determine the charge on the capacitor. Q0 = Determine the energy stored in the capacitor. PE0 = 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? E = What are the new value of the capacitance? C = What are the new value of thethe charge on the capacitor. Q = What are the new value the energy stored in the capacitor? PE =

Explanation / Answer

eletric field E = V/d

E = 35/ 3mm

E = 1.166 *10^4 V/m

b. Capacitance C = eoA/d

= 8.85*10^-12 *2.5/3mm

= 7.375 nF

c. charge Q = CV

Q = 7.375 nF * 35

Q = 2.581*10^-7 C or 0.258 uC

energy stored U = 0.5 cv^2

U = 0.5 * 7.375 *10^-9 *35*35

U = 4.517 *10^-6 Joules

e. EFnew = Eo/K

= 1.166*10^4/3.2

= 3.64 *10^3 V/m

f. Cnew = Kco

Cnew = 3.2* 7.375 nF

Cnew = 23.6 nF

g. charge new = CV

= 23.6 * 35

= 8.26 *10^-7 C or 0.826 uC

h. PE new = 0.5 cv^2 = 0.5 * 23.6nF * 35*35

PEnew = 14.455 uJoules

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