Two parallel plates, each having area A = 2002 cm 2 are connected to the termina

Question

Two parallel plates, each having area A = 2002 cm2are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.38 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 ( = 0.76 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

Explanation / Answer

(1)

, the charge on the top plate is

Q= A eo / d * V

=2002* 10^-4 m^2 ( 8.85 * 10^-12/0.38 * 10^-2 * 6

=2.79 * 10^-9 C

(2)

the energy stored in the this capacitor is

E = QV/2

= 2.79 * 10^-9 C(6)/2

=8.37 * 10^-9 J

(3)

Since the battery is first disconnected, it can do no work and so the charge on the plates remains
constant. However moving the plate apart does work and changes the voltage. We saw above that the
voltage is directly proportional to the distance. So doubling the distance doubles the voltage. From
above we saw that energy is given by:

So the net effect is that the energy doubles.

E = 16.74 * 10^-9 J

(4)

E = sigma/ eo = Q/Aeo = 2.79 * 10^-9 C/2002* 10^-4 (8.85 * 10^-12)

=1574.69 N/C

(5)

V>Vb

Like we showed in (3) doubling the distance while the charge stays the same means the potential
difference doubles from what it was before.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.