Two parallel plates, each having area A = 3611 cm 2 are connected to the termina

Question

Two parallel plates, each having area A = 3611 cm2are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.56 cm.

1. What is Q, the charge on the top plate?

2. What is U, the energy stored in the this capacitor?

3. The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.12 cm). What is the energy stored in this new capacitor?

4. What is E, the magnitude of the electric field in the region between the plates?

Explanation / Answer

We know the Q = CV

C = epsilon*A/d = 8.85 x 10-12 * 0.3611 / 0.56x10-2 =5.7066 x 10-10 F

Q = 5.7066 x 10-10 F *6.0V = 3.424 x 10-9 Coulomb .......Ans.

2. Energy = 1/2 C*V2 =0.5* 5.7066 x 10-10 * 36 = 1.027188 x 10-8 J......Ans.

3. If the distance is doubled then C =5.7066 x 10-10 F / 2 = 2.8533 x 10-10 F

hence Energy will also halfed = 0.513 x 10-8 J......Ans.

4. E = kQ/r = 9x109 * 3.424 x 10-9 / 0.56x10-2 =5502.85 N/C

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