Two parallel plates, each having area A = 2577 cm 2 are connected to the termina

Question

Two parallel plates, each having area A = 2577 cm2are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.55 cm.

1)

What is Q, the charge on the top plate?C

2)

What is U, the energy stored in the this capacitor?J

3)

The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.1 cm). What is the energy stored in this new capacitor?J

4)

What is E, the magnitude of the electric field in the region between the plates?N/C

5)

Compare V, the magnitude of the new potential difference across the plates, to Vb, the voltage of the battery.

V < Vb

V = Vb

V > Vb

6)

Two uncharged parallel plates are now connected to the initial pair of plates as shown. How will the electric field, E, and potential difference across the plates, V, change, if at all?

Both E and V will remain the same

E will decrease and V will increase

E will increase and V will decrease

Both E and V will decrease

Both E and V will increase

Explanation / Answer

First of all we have to calculate the value of capacitance
C = eoA/d where C is the capacitance , A is the area and d is the distance.
C = (8.85*10-12*2577*10-4)/(0.55*10-2) = 41.466*10-11 F
(1) Charge Q = CV = 41.466*10-11*6= 248.79*10-11 C
(2) Energy = (1/2)CV2 = (1/2) 41.466*10-11*62 = 746.388*10-11 J
(3) If battery is disconnected then the process will continue on constant charge
therfore energy = Q2/2C
and C becomes half as of earlier
so new C = (1/2)*41.466*10-11 F
Energy = (248.79*10-11 )2/(2*(1/2)*41.466*10-11 ) = 1492.70*10-11 J
(4) Electric field E = Q/Aeo = 248.79*10-11 /(2577*10-4*8.85*10-12) = 0.10908 N/C

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