Question

Direction. Explain the following.

1. What is the relationship between the electric field E and the electric potential

V between the plates of the capacitor? Explain.

2. Where is the energy stored in a parallel-plate capacitor? Explain.

3. Let the energy stored “in the capacitor” be U. Show that U is given by the

expression: U = 1

2

Q2

/C

[Hint: The power P in the capacitor is given by P = 𝑑𝑈

𝑑𝑡

= IV where I = 𝑑𝑄

𝑑𝑡

. This is

a simple exercise on integration.]

Answer 1

1) The relationship between V and E for parallel conducting plates is

$$E=\frac{V}{d}$$

2) It is between the capacitor plates. A charged capacitor stores energy in the electrical field between its plates. As the capacitor is being charged, the electrical field builds up. When a charged capacitor is disconnected from a battery, its energy remains in the field in the space between its plates.

3)

$$P=\frac{dU}{dt}=VI\\P=\frac{dU}{dt}=V\frac{dq}{dt}\\ dU=Vdq\\dU=\frac{q}{C}dq\\U=\frac{q^2}{2C}$$