Two parallel plates, each having area A = 2826 cm 2 are connected to the termina

Question

Two parallel plates, each having area A = 2826 cm2are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.33 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 ( = 0.66 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

given that

area A = 2826 cm^2 = 0.2826 m^2

battery of voltage Vb = 6 V .

plates are separated by a distance d = 0.33 cm = 0.0033 m

we know that

C =e0*A/d

C = 8.85*10^(-12)*0.2826 / 0.0033

C = 2.50*10^(-12) / 0.0033

C = 757.88 *10^(-12) F

C = 7.57*10^(-10) F

part(1)

Q = C*Vb

Q = 7.57*10^(-10) *6

Q = 4.54*10^(-9) C

part(2)

U = 1/2*C*Vb^2

U = 1/2*7.57*10^(-10)*(6^2)

U = 136.26 *10^(-10)

U = 1.36*10^(-8) J

part(3)

we know

Q = C*V

and C =e0*A/d

when d is double C should be half.

and Q remains constant .

V = Q/C

so when C is half than V shold be double .

now voltage V = 12 V

U =1/2*C/2*V^2

now new potential energy

U =1/2*(7.57*10^(-10) /2) *12^2

U = 2.72*10^(-8) J

part(4)

E = V/d

E = 12/0.0033

E = 3636.36 V/m

part(5)

new potential difference across the plates V is double of voltage across the battery Vb.

so V > Vb

part(6)

from the figure we can see that both parallel plate capacitors are in parallel cobination

so equavalent capacitance is Ceq.

distance b/w plates is doubled so capacitance is to be half .

so Ceq = C/2 + C/2 = C, it means C remains constant for this combination .

and Q also remain constant so

V and E also remain the same .

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