Two parallel plates, each having area A = 2713 cm 2 are connected to the termina

Question

Two parallel plates, each having area A = 2713 cm2are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.59 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 ( = 1.18 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

Explanation / Answer

We know that capacitance will be

AE0/d

where A is the area of the plates, and d is the sepration between the plates

E0= 8.86*10-12

So initially

C= 4.074*10-10F

Now we know that,

Q=CV

so ,

Q=6* 4.074*10-10

Q=2.438 *10-9C

2)

We know that

U=CV2/2

So

U=7.3332*10-9J

3)

Now as the distance between the plates is doubled so capactiance will be reduced to half, as it is inversly proptional to the separation between the plates.

ANd charge will remain same when the battery is disconnected so ,

U=Q2/2C..................C=2.037*10-10F

U=2.01545*10-9

4.

Now we know that E=Q/AE0

E=1015.408131 N/C

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