# 4.10 Electrostatic Potential Energy

A source connected to a capacitor expends energy in charging up the capacitor. If the capacitor plates are made of a good conductor with effectively zero resistance, and if the dielectric separating the two plates has negligible conductivity, then no real current can flow through the dielectric, and no ohmic losses occur anywhere in the capacitor. Where then does the energy expended in charging up the capacitor go? The energy ends up getting stored in the dielectric medium in the form of electrostatic potential energy. The amount of stored energy W e is related to Q, C, and V.

Suppose we were to charge up a capacitor by ramping up the voltage across it from υ=0 to υ=V. During the process, charge +q accumulates on one conductor and −q on the other. In effect, a charge q has been transferred from one of the conductors to the other. The voltage υ across the capacitor is related to q by

υ=q C.(4.128)

From the definition of υ, the amount of work d W e required to transfer an additional incremental charge d q from one conductor to the other is

d W e=υ d q=q C d q.(4.129)

If we transfer a total charge Q between the conductors of an initially uncharged capacitor, then the total amount of work performed is

W e=∫0 Q q C d q=1 2 Q 2 C(J).(4.130)

Using C=Q/V, where V is the final voltage, W e also can be expressed as

The capacitance of the parallel-plate capacitor discussed in Example 4-14 is given by Eq. (4.123) as C=ε A/d, where A is the surface area of each of its plates and d is the separation between them. Also, the voltage V across the capacitor is related to the magnitude of the electric field E in the dielectric by V=E d. Using these two expressions in Eq. (4.131) gives

W e=1 2 ε A d(E d)2=1 2 ε E 2(A d)=1 2 ε E 2 υ,(4.132)

where υ=A d is the volume of the capacitor. This expression affirms the assertion made at the beginning of this section, namely that the energy expended in charging up the capacitor is being stored in the electric field present in the dielectric material in between the two conductors.

The electrostatic energy density w e is defined as the electrostatic potential energy W e per unit volume:

Even though this expression was derived for a parallel-plate capacitor, it is equally valid for any dielectric medium containing an electric field E, including a vacuum. Furthermore, for any volume υ, the total electrostatic potential energy stored in it is

W e=1 2∫υ ε E 2 d υ(J).(4.134)

Returning to the parallel-plate capacitor, the oppositely charged plates are attracted to each other by an electrical force F e. In terms of the coordinate system of Fig. 4-28, the electrical force acting on the upper plate is along −z ˆ (due to attraction by the lower plate). Hence, it is given by

F e=−z ˆ F e(force on upper plate).(4.135)

# Figure 4-28 A dc voltage source connected to a parallel-plate capacitor.

Figure 4 28 Full Alternative Text

Our plan is to compute F e from energy considerations. We start by converting the spacing d into a variable z and using C=ε A/z in Eq. (4.131):

W e=1 2 C V 2=1 2 ε A V 2 z.(4.136)

If V is maintained at a constant level, W e decreases when increasing the separation z between the plates. If an external, upward-directed force F=−F e is applied to counter the electrostatic force F e and used to move the upper plate upwards by a distance d z, the expended mechanical work is

d W=F⋅z ˆ d z.(4.137)

The work d W is equal to the loss in electrostatic energy stored in the capacitor. That is,

d W=−d W e.(4.138)

Also, F e=−F, which leads to

d W e=F e⋅z ˆ d z=−z ˆ F e⋅z ˆ d z=−F e d z.(4.139)

From Eq. (4.136),

d W e=−1 2 ε A V 2 z 2 d z.(4.140)

Equating Eqs. (4.139) and (4.140) and replacing z with d leads to

F e=1 2 ε A V 2 d 2, (4.141a)

and

This is the electrostatic force exerted on the upper plate. The force on the lower plate is identical in magnitude and opposite in direction.

The relationship given by Eq. (4.139) pertains to a capacitor with d l=z ˆ d z. We can generalize the result for d l along any direction as

# Example 4-16: Force on Sliding Dielectric

The two plates of the parallel-plate capacitor shown in Fig. 4-29 are each of length ℓ and width w, and the separation between them is d. The capacitor contains a dielectric block of dimensions ℓ×w×d and permittivity ε. The block can slide in and out of the capacitor cavity along its length dimension. Compute the force F e acting on the dielectric block when it is partially outside of the cavity and the voltage across the capacitor is V.

# Figure 4-29 Parallel-plate capacitor with slidable dielectric block.

Figure 4 29 Full Alternative Text

Solution: From Eq. (4.122), the electric field inside the capacitor cavity is

E=V d.

This is true in both the section containing the dielectric block and the section filled with air. The total electrostatic energy of the capacitor consists of two components: one for the volume containing the dielectric block of permittivity ε and volume υ 1=x w d, and another for the volume containing air with ε 0 and volume υ 2=(ℓ−x)w d. Hence,

W e=1 2 ε E 2 υ 1+1 2 ε 0 E 2 υ 2=1 2 ε V d 2 x w d+1 2 ε 0 V d 2(ℓ−x)w d=1 2 V 2 d w[ε x+ε 0(ℓ−x)].(4.143)

Since ε>ε 0, the electrostatic energy is maximum when x=ℓ (dielectric block fully inside the cavity). Sliding the dielectric block out of the capacitor requires exerting an external mechanical force F to oppose the electrostatic force F e, whose tendency is to oppose reduction in W e. Thus, the direction of F e is to pull the block back into the capacitor.

The magnitude of F e can be obtained from

F e=d W e d x=d d x 1 2 V 2 d w[ε x+ε 0(ℓ−x)]=1 2 V 2 d w(ε−ε 0).(4.144)

Concept Question 4-26: To bring a charge q from infinity to a given point in space, a certain amount of work W is expended. Where does the energy corresponding to W go?

Concept Question 4-27: When a voltage source is connected across a capacitor, what is the direction of the electrical force acting on its two conducting surfaces?