Electric Potential Energy with Several Point Charges

Suppose the electric field $\mathbf{E}$ in which charge $q_0$, moves is caused by several point charges $q_1$, $q_2$, $q_3$, ... from $q_0$ at distances $r_1$, $r_2$, $r_3$, ... from $q_0$ as in Fig.1. For example, $q_0$ could be a positive ion moving in the presence of other ions (Fig. 2). The total electric field at each point is the vector sum of the fields due to the individ ual charges, and the total work done on $q_0$ during any displacement is the sum of the contributions from the individual charges. 

Fig.1: The potential energy associated with a charge $q_0$ at point a depends on the other charges $q_1$, $q_2$, and $q_3$ and on their distances $r_1$, $r_2$ and $r_3$ from point a.

From Eq. (2) we conclude that the potential energy associated with the test charge $q_0$ at point a in Fig. 1 is the algebraic sum (not a vector sum):

$U=\frac{q_0}{4\pi \epsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}+...\right)=\frac{q_0}{4\pi \epsilon_0}\Sigma_i \frac{q_i}{r_i}$ (point charge $q_0$ and collection of charges $q_i$)

When $q_0$ is at a different point b the potential energy is given by the same expression, but $r_1$, $r_2$, ... are the distances from $q_1$, $q_2$, . . . to point b. The work done on charge $q_0$ when it moves from a to b along any path is equal to the difference $U_a-U_b$ between the potential energies when $q_0$ is at a and at b.

We can represent any charge distribution as a collection of point charges, so Eq. (23.10) shows that we can always find a potential-energy function for any static electric field. It follows that for every electric field due to a static charge distribution, the force exerted by that field is conservative.

Equations (23.9) and (23.10) define U to be zero when all the distances $r_1$, $r_2$, . . . are infinite—that is, when the test charge $q_0$ is very far away from all the charges that produce the field. As with any potential-energy function, the point where $U=0$ is arbitrary; we can always add a constant to make U equal zero at any point we choose. In electrostatics problems it’s usually simplest to choose this point to be at infinity. When we analyze electric circuits in Chapters 25 and 26, other choices will be more convenient. 

Equation (23.10) gives the potential energy associated with the presence of the test charge in the field produced by But there is also potential energy involved in assembling these charges. If we start with charges all separated from each other by infinite distances and then bring them together so that the distance between and is the total potential energy U is the sum of the potential energies of interaction for each pair of charges. We can write this as

$U=\frac{1}{4\pi \epsilon}\Sigma_{i<j}\frac{q_iq_j}{r_{ij}}$

This sum extends over all pairs of charges; we don’t let $i=j$ (because that would be an interaction of a charge with itself ), and we include only terms with $i<j$ to make sure that we count each pair only once. Thus, to account for the interaction between $q_3$ and $q_4$, we include a term with i = 3 and j = 4 but not a term with i = 4 and j = 3.

Interpreting Electric Potential Energy

As a final comment, here are two viewpoints on electric potential energy. We have defined it in terms of the work done by the electric field on a charged particle moving in the field, just as in Chapter 7 we defined potential energy in terms of the work done by gravity or by a spring. When a particle moves from point a to point b the work done on it by the electric field is $W_{a→b}=U_a-U_b$. Thus the potential-energy difference $U_a-U_b$ equals the work that is done by the electric force when the particle moves from a to b. When $U_a$ is greater than $U_b$, the field does positive work on the particle as it “falls” from a point of higher potential energy (a) to a point of lower potential energy (b).

An alternative but equivalent viewpoint is to consider how much work we would have to do to “raise” a particle from a point b where the potential energy is $U_b$ to a point a where it has a greater value $U_a$ (pushing two positive charges closer together, for example). To move the particle slowly (so as not to give it any kinetic energy), we need to exert an additional external force $\mathbf{F_{ext}}$ that is equal and opposite to the electric-field force and does positive work. The potential-energy difference $U_a-U_b$ is then defined as the work that must be done by an external force to move the particle slowly from b to a against the electric force.

Because $\mathbf{F_{ext}}$ is the negative of the electric-field force and the displacement is in the opposite direction, this definition of the potential difference $U_a-U_b$ is equiv-alent to that given above. This alternative viewpoint also works if $U_a$ is less than $U_b$ corresponding to “lowering” the particle; an example is moving two positive charges away from each other. In this case, $U_a-U_b$ is again equal to the work done by the external force, but now this work is negative.

We will use both of these viewpoints in the next section to interpret what is meant by electric potential, or potential energy per unit charge.

Example 1

Four charges are arranged at the corners of a square ABCD of side d, as shown in Fig. 2.15.(a) Find the work required to put together this arrangement. (b) A charge $q_0$ is brought to the centre E of the square, the four charges being held fixed at its corners. How much extra work is needed to do this?

Fig.2

Solution:

(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose,

first the charge $+q$ is brought to A, and then the charges $–q$, $+q$, and $–q$ are brought to B, C and D, respectively. The total work needed can be calculated in steps:

(i) Work needed to bring charge $+q$ to A when no charge is present elsewhere: this is zero.

(ii) Work needed to bring $–q$ to B when $+q$ is at A. This is given by (charge at B) × (electrostatic potential at B due to charge $+q$ at A)

$=-q \times \left(\frac{q}{4\pi \epsilon_0d}\right)=-\frac{q^2}{4\pi \epsilon_0 d}$

(iii) Work needed to bring charge $+q$ to C when $+q$ is at A and $–q$ is at B. This is given by (charge at C) × (potential at C due to charges at A and B)

$=+q \left(\frac{+q}{4\pi \epsilon_0d \sqrt{2}}+\frac{-q}{4\pi \epsilon_0d}\right)=-\frac{q^2}{4\pi \epsilon_0 d}$

$=\frac{-q^2}{4\pi \epsilon_0 d}\left(1-\frac{1}{\sqrt{2}}\right)$

(iv) Work needed to bring $–q$ to D when $+q$ at A, $–q$ at B, and $+q$ at C. This is given by (charge at D) × (potential at D due to charges at A, B and C)

$=-q \left(\frac{-q}{4\pi \epsilon_0d }+\frac{q}{4\pi \epsilon_0d\sqrt{2}}+\frac{q}{4\pi \epsilon_0d }\right)=-\frac{q^2}{4\pi \epsilon_0 d}$

$=\frac{-q^2}{4\pi \epsilon_0 d}\left(2-\frac{1}{\sqrt{2}}\right)$

Add the work done in steps (i), (ii), (iii) and (iv). The total work required is

$=\frac{-q^2}{4\pi \epsilon_0d}\left[(0)+(1)+\left(1-\frac{1}{\sqrt{2}}\right)+\left(2-\frac{1}{\sqrt{2}}\right)\right]$

$=\frac{-q^2}{4\pi \epsilon_0 d}(4-\sqrt{2})$

The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges. (Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy

will remain the same.)

(b) The extra work necessary to bring a charge $q_0$ to the point E when the four charges are at A, B, C and D is $q_0$ × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. Hence, no work is required to bring any charge to point E.

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