2.8 Potential Energy in an External Field

2.8.1 Potential energy of a single charge

In Section 2.7, the source of the electric field was specified, i.e., the charges and their locations, and the potential energy of the system of those charges could be determined.

In this section, we ask a related but a distinct question. What is the potential energy of a charge q in a given external field? This question was, in fact, the starting point that led us to the notion of the electrostatic potential (Sections 2.1 and 2.2). But here we address this question again to clarify in what way it is different from the discussion in Section 2.7.

The main difference is that we are now concerned with the potential energy of a charge (or charges) in an external field. The external field E vector is not produced by the given charge(s) whose potential energy we wish to calculate. E vector is produced by sources external to the given charge(s).The external sources may be known, but often they are unknown or unspecified; what is specified is the electric field E vector or the electrostatic potential V due to the external sources. We assume that the charge q does not significantly affect the sources producing the external field. This is true if q is very small, or the external sources are held fixed by other unspecified forces. Even if q is finite, its influence on the external sources may still be ignored in the situation when very strong sources far away at infinity produce a finite field E vector in the region of interest. Note again that we are interested in determining the potential energy of a given charge q (and later, a system of charges) in the external field; we are not interested in the potential energy of the sources producing the external electric field.

The external electric field E vector and the corresponding external potential V may vary from point to point. By definition, V at a point P is the work done in bringing a unit positive charge from infinity to the point P.

(We continue to take potential at infinity to be zero.) Thus, work done in bringing a charge q from infinity to the point P in the external field is qV. This work is stored in the form of potential energy of q. If the point P has position vector r relative to some origin, we can write:

Potential energy of q at r vector in an external field


= qV(r vector) .. .. (2.27)

where V(r vector) is the external potential at the point r vector.

Thus, if an electron with charge q = e = 1.6 × 10^−19 C is accelerated by a potential difference of Δ V = 1 volt, it would gain energy of q ΔV = 1.6 × 10^−19J. This unit of energy is defined as 1 electron volt or 1eV, i.e., 1 eV = 1.6 × 10^−19J.


The units based on eV are most commonly used in atomic, nuclear and particle physics,


(1 k eV = 10^3 eV = 1.6 × 10^−16 J,


1 M eV = 10^6 eV = 1.6 × 10^−13 J,


1 GeV = 10^9 eV = 1.6 × 10^−10 J and


1 TeV = 10^12 eV = 1.6 × 10^−7 J).


[This has already been defined on Page 117, class 11 Physics Part 1, Table 6.1.]

2.8.2 Potential energy of a system of two charges in an external field

Next, we ask: what is the potential energy of a system of two charges q1 and q2 located at r1 vector and r2 vector, respectively, in an external field? First, we calculate the work done in bringing the charge q1 from infinity to r1 vector. Work done in this step is q1 V (r1 vector), using Equation (2.27). Next, we consider the work done in bringing q2 to r2 vector. In this step, work is done not only against the external field E vector but also against the field due to q1.

Work done on q2 against the external field


= q2 V (r2 vector)

Work done on q2 against the field due to q1


= q1q2/4πε0 r12

where r12 is the distance between q1 and q2.


We have made use of Equations (2.27) and (2.22).


By the superposition principle for fields, we add up the work done on q2 against the two fields (E vector and that due to q1): Work done in bringing q2 to r2 vector

= q2V (r2 vector) + q1q2/4πε0 r12 .. .. (2.28)

Thus,


Potential energy of the system


= the total work done in assembling the configuration


= q1V (r1 vector) + q2V (r2 vector) + q1q2/4πε0 r12 .. .. (2.29)

Example 2.5

(a) Determine the electrostatic potential energy of a system consisting of two charges 7 μC and −2 μC (and with no external field) placed at (−9 cm, 0, 0) and (9 cm, 0, 0) respectively.


(b) How much work is required to separate the two charges infinitely away from each other?


(c) Suppose that the same system of charges is now placed in an external electric field E = A (1/r^2); A = 9 × 10^5 C m^−2. What would the electrostatic energy of the configuration be?

Solution:


(a) U = 1/4πε0 × q1q2/r = 9 × 10^9 × (7 × (−2) × 10^−12)/0.18


= −0.7 J.

(b) W = U2 − U1


= 0 − U = 0 − (−0.7) = 0.7 J.

(c) The mutual interaction energy of the two charges remains unchanged. In addition, there is the energy of interaction of the two charges with the external electric field. We find,


q1 V (r1 vector) + q2 V (r2 vector) = A × 7 μC/0.09 m + A × −2μC/0.09 m

and the net electrostatic energy is

q1 V (r1 vector) + q2 V (r2 vector) + q1q2/4πε0 r12


= A × (7 μC/0.09 m) + A × (−2μC/0.09 m) − 0.7 J


= 70 − 20 − 0.7 = 49.3 J

2.8.3 Potential energy of a dipole in an external field

Consider a dipole with charges q1 = +q and q2 = −q placed in a uniform electric field E vector, as shown in Fig.2.16.

FIGURE 2.16 Potential energy of a dipole in a uniform external field.

As seen in the last chapter, in a uniform electric field, the dipole experiences no net force; but experiences a torque τ vector given by


τ vector = p vector ⨯ E vector .. .. (2.30)

which will tend to rotate it (unless p vector is parallel or antiparallel to E vector). Suppose an external torque τ ext is applied in such a manner that it just neutralises this torque and rotates it in the plane of paper from angle θ0 to angle θ1 at an infinitesimal angular speed and without angular acceleration. The amount of work done by the external torque will be given by


W = ∫ from θ0 to θ1 (τ ext (θ) dθ)

= pE (cos θ0 − cos θ1).. .. (2.31)

This work is stored as the potential energy of the system. We can then associate potential energy U (θ) with an inclination θ of the dipole. Similar to other potential energies, there is a freedom in choosing the angle where the potential energy U is taken to be zero. A natural choice is to take θ0 = π/2.


(An explanation for it is provided towards the end of discussion.)

We can then write,


U (θ) = pE (cos π/2 − cos θ)


= p E cosθ


= − p vector⋅ E vector .. .. (2.32)

This expression can alternately be understood also from Equation (2.29).


We apply Equation (2.29) to the present system of two charges +q and −q. The potential energy expression then reads

U′ (θ) = q [V (r1 vector) − V (r2 vector)] − {q^2/ (4πε0 × 2 a)} .. .. (2.33)

Here, r1 vector and r2 vector denote the position vectors of +q and −q. Now, the potential difference between positions r1 vector and r2 vector equals the work done in bringing a unit positive charge against field from r2 vector to r1 vector. The displacement parallel to the force is 2a cos θ. Thus,


[V (r1 vector) − V (r2 vector)] = −E × 2a cos θ.

We thus obtain,


U′ (θ) = −pE cos θ − {q^2/ (4πε0 × 2 a)}

= −p vector ⋅ E vector − {q^2/ (4πε0 × 2 a)} .. .. (2.34)

We note that U′ (θ) differs from U (θ) by a quantity which is just a constant for a given dipole. Since a constant is insignificant for potential energy, we can drop the second term in Equation (2.34) and it then reduces to Equation (2.32).

We can now understand why we took θ0 = π/2. In this case, the work done against the external field E vector in bringing +q and −q are equal and opposite and cancel out, i.e., q [V (r1 vector) − V (r2 vector)] = 0.

Example 2.6

A molecule of a substance has a permanent electric dipole moment of magnitude 10^−29 C m. A mole of this substance is polarized (at low temperature) by applying a strong electrostatic field of magnitude 10^6 V m^−1. The direction of the field is suddenly changed by an angle of 60 °. Estimate the heat released by the substance in aligning its dipoles along the new direction of the field. For simplicity, assume 100% polarization of the sample.

Solution:


Here, dipole moment of each molecules = 10^−29 C m

As 1 mole of the substance contains 6 × 10^23 molecules,


total dipole moment of all the molecules, p


= 6 × 10^23 × 10^−29 C m


= 6 × 10^−6C m

Initial potential energy, Ui = −pE cos θ = −6 × 10^−6 × 10^6 cos 0° = −6 J

Final potential energy (when θ = 60°), Uf = −6 × 10^−6 × 10^6 cos 60° = −3 J

Change in potential energy = −3 J − (−6J) = 3 J

So, there is loss in potential energy. This must be the energy released by the substance in the form of heat in aligning its dipoles.