Answer to Question #135880 in Electric Circuits for Afraid

Question

Explain energy of a capacitor?

Answer 1

The capacitor energy is the potential energy of the interaction of the capacitor plates - after all, the plates, being charged differently, are attracted to each other.

Take such a small platform on the second plate of the capacitor that the charge $q_0$ this site can be considered point. This charge is attracted to the first lining with force $F_0 = q_0E_1$ , where $E_1$ is the field strength of the first lining:

$E_1=\frac{\displaystyle \sigma }{\displaystyle 2 \varepsilon _0 \vphantom{1^a}}=\frac{\displaystyle q}{\displaystyle 2\varepsilon_0 S \vphantom{1^a}}.$

Therefore,

$F_0=\frac{\displaystyle q_0q}{\displaystyle 2 \varepsilon_0 S \vphantom{1^a}}.$

This force is directed parallel to the field lines (i.e. perpendicular to the plates).

The resulting attraction force $F$ of the second cover to the first cover is composed of all these forces $F_0$ with which all kinds of small charges $q_0$ the second cover are attracted to the first cover. In this summation, the constant multiplier $\frac q {2\varepsilon _ 0 S}$ is placed outside the bracket, and in the bracket all $q_0$ are summed and $q$ is given. As a result, we get:

$F=\frac{\displaystyle q^2}{\displaystyle 2 \varepsilon_0 S \vphantom{1^a}}$ . (1)

Suppose now that the distance between the plates has changed from the initial value $d_1$ to the final value $d_2$ . The force of attraction of the plates performs the following work:

$A = F(d_1 - d_2)$ .

The sign is correct: if the plates approach $(d_2 < d_1)$ , then the force performs positive work, since the plates are attracted to each other. On the contrary, if you remove the plates $(d_2 > d_1)$ , then the operation of the attraction force is negative, as it should be.

Taking into account the formulas (1) and $C=\frac{\displaystyle \varepsilon_0 S}{\displaystyle d \vphantom{1^a}}$ - formula of flat air condenser capacity, we have:

$A=\frac{\displaystyle q^2}{\displaystyle 2 \varepsilon_0 S \vphantom{1^a}}\left ( d_1-d_2 \right )=\frac{\displaystyle q^2d_1}{\displaystyle 2\varepsilon_0 S \vphantom{1^a}}-\frac{\displaystyle q^2d_2}{\displaystyle 2\varepsilon_0 S \vphantom{1^a}}$ $=\frac{\displaystyle q^2}{\displaystyle 2C_1 \vphantom{1^a}}-\frac{\displaystyle q^2}{\displaystyle 2C_2 \vphantom{1^a}}=W_1-W_2,$

where

$W_1=\frac{\displaystyle q^2}{\displaystyle 2C_1 \vphantom{1^a}},$

$W_2=\frac{\displaystyle q^2}{\displaystyle 2C_2 \vphantom{1^a}}$

This can be rewritten as follows:

$A = -(W_2 - W_1) = - \Delta W,$

where

$W=\frac{\displaystyle q^2}{\displaystyle 2C \vphantom{1^a}}$ .

The operation of the potential attraction force $F$ of the plates turned out to be equal to a change with a sign minus the value of $W$. This just means that $W$ is the potential interaction energy of the plates, or the energy of a charged capacitor.