Two parallel plates, each having area A = 2263 cm 2 are connected to the termina

Question

Two parallel plates, each having area A = 2263 cm2are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.46 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 ( = 0.92 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

6Two 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

here,

area , A= 2263cm^2

A = 0.2263 m^2

Vb = 6 V

d = 0.46 cm

d = 0.0046 m

capacitance , C= A*epsilon0/d

C = 0.2263 * 8.85*10^-12/0.0046

C = 4.35 * 10^-10 F

(a)

the charge stored , Q = C*V

Q = 2.61 * 10^-9 C

the charge stored is 2.61 * 10^-9 C

(b)

the energy stored in the capacitor , E = 0.5 * C * V^2

E = 0.5 * 4.35 * 10^-10 * 6^2

E = 7.83 * 10^-9 J

the energy stored in the capacitor is 7.83 * 10^-9 J

(c)

d' = 0.92 cm

d' = 0.0092 m

capacitance , C' = A*epsilon0/d

C' = 0.2263 * 8.85*10^-12/0.0096

C' = 2.09 * 10^-10 F

the energy stored in the capacitor , E = 0.5 *Q^2/C

E = 0.5 * ( 2.61 * 10^-9 )^2 /(2.09 * 10^-10 )

E = 1.63 * 10^-8 J

the energy stored in the capacitor is 1.63 * 10^-8 J

(d)

V' = Q/C'

V' = 12.49 V

electric feild , E = V/d

E = 12.49 / 0.0096

E = 1300.84 N/C

the electric feild is 1300.84 N/C

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