Potential Energy of a System of Charges

Consider two charges q₁ and q₂ placed at position vectors r₁ and r₂.

To find the potential energy of the system, we calculate the work done by an external agent in bringing the charges from infinity to their positions. First, charge q₁ is brought from infinity to position r₁. Since no other charge is present, no electric field opposes its motion, so the work done is zero. Once placed at r₁, charge q₁ creates an electric potential at any point P given by

 V_1=\frac{1}{4\pi\epsilon_0}\frac{q_1}{r_{1P}}

Let r₁P be the distance of a point P from charge q₁. By definition of potential, the work done in bringing q₂ from infinity to r₂ is

\text{work done on q}_2=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}} 

where r₁₂ is the distance between q₁ and q₂. Since electrostatic forces are conservative, this work is stored as the potential energy of the system:

U=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}} ..... (1)                                                                                              

The order of placing the charges does not affect U.

• U > 0 when q₁q₂ > 0 (like charges repel; work must be done to bring them together).
• U < 0 when q₁q₂ < 0 (opposite charges attract; work is done by the field in bringing them together, so external work is negative).

Potential energy reflects the nature of the electrostatic interaction between the charges.

Equation (1) can be generalised to a system of any number of point charges. Consider three charges q₁, q₂​, and q₃ at positions r₁, r₂​, and r₃​. Bringing q₁​ from infinity to r₁ requires no work. Next, bringing q₂​ from infinity to r₂​ requires work equal to q₂ times the potential at r₂​ due to q₁,

q_2V_1(r_2)=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}}   .....(2)     

The charges q₁ and q₂ generate a potential, which at any point P can be written as

V_{1,2}=\frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{r_{1P}}+\frac{q_2}{r_{2P}}\right)

Work done next in bringing q₃ from infinity to the point r₃ is q₃ times V_1,2 at r₃ can be written as,

q_3 V_{1,2}(r_3) = \frac{1}{4 \pi \varepsilon_0} \left( \frac{q_1 q_3}{r_{13}} + \frac{q_2 q_3}{r_{23}} \right) ....(3)

The total work done in collecting the charges at the given locations is obtained by adding equations (2) and (3),

U=\frac{1}{4\pi\epsilon_0}\left(\frac{q_1q_2}{r_{12}}+\frac{q_1q_3}{r_{13}}+\frac{q_2q_3}{r_{23}}\right)

Because electrostatic forces are conservative, the potential energy depends only on the configuration of charges, not on the order in which they were assembled.

Potential due to a System of Charges

Suppose a system of charges q₁, q₂,…, q_n with position vectors r₁, r₂,…, r_n relative to some origin. The potential V₁ at P due to the charge q₁ is

V_1=\frac{1}{4\pi\epsilon_0}\frac{q_1}{r_{1P}}

where r1P is the distance between q1 and P.

Similarly, the potential V₂ at P due to q₂ and V₃ due to q₃ can be written as,

V_2=\frac{1}{4\pi\epsilon_0}\frac{q_2}{r_{2P}}\\ V_3=\frac{1}{4\pi\epsilon_0}\frac{q_3}{r_{3P}}

where r_2P and r_3P are the distances of P from charges q₂   ​and q₃ ​, respectively; and so on for the potential due to other charges. By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charge that is,

V = V₁ + V₂ + V₃ +...... + V_n

The above expression can be expressed as,

V=\frac{1}{4\pi\epsilon_0}\frac{q_1}{r_{1P}}+\frac{1}{4\pi\epsilon_0}\frac{q_2}{r_{2P}}+ \frac{1}{4\pi\epsilon_0}\frac{q_3}{r_{3P}}+.....+\frac{1}{4\pi\epsilon_0}\frac{q_n}{r_{nP}}\\ V=\frac{1}{4\pi\epsilon_0}\left(\frac{q_1}{r_{1P}}+\frac{q_2}{r_{2P}}+\frac{q_3}{r_{3P}}+...+\frac{q_n}{r_{nP}}\right)

A continuous charge distribution with a charge density ρ (r), must be divided into small volume elements of size ∆v, each carrying a charge ρ ∆v. Then calculate the potential due to each volume element and add (or, more precisely, integrate) all of these contributions to get the total potential due to the distribution.

Sample Problems  

Question 1: Two charges 3 × 10^–8 C and –2 × 10^–8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. 

Solution: Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin.

Let P be the expected point on the x-axis where the potential is zero. If x is the x-coordinate of P, and therefore x must be positive. If x lies between O and A, then

\frac{1}{4\pi\epsilon_0}\left[\frac{3\times10^{-8}}{x\times10^{-2}}-\frac{2\times10^{-8}}{(15-x)\times10^{-2}}\right]=0

Rearrange the above equation to find the value of x,

\frac{3}{x}-\frac{2}{(15-x)}=0\\ 45-3x-2x=0\\ 5x=45\\ x=9

which gives x = 9 cm.

If x lies on the extended line OA, the required condition is

\frac{3}{x}-\frac{2}{(x-15)}=0\\ 3x-45-2x=0\\ x=45

which results x = 45 cm.

On the side of the negative charge, an electric potential is zero at 9 cm and 45 cm away from the positive charge.

Question 2: Figures (a) and (b) show the field lines of a positive and negative point charge respectively.

(a) Give the signs of the potential difference V_P – V_Q; V_B – V_A.

(b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B.

(c) Give the sign of the work done by the field in moving a small positive charge from Q to P.

(d) Give the sign of the work done by the external agency in moving a small negative charge from B to A.

(e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A? 

Solution:

(a) As V ∝ (1/r) and V_P> V_Q. Thus, (V_P – V_Q) is positive. Also, V_B is less negative than V_A. Thus, V_B > V_A or (V_B – V_A) is positive.

(b) A tiny negative charge will be attracted towards a positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore, the sign of potential energy difference of a small negative charge between Q and P is positive. Similarly, V_A > V_B and hence the sign of potential energy differences are positive.

(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.

(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.

(e) Due to the force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.

Question 3: A 500 µC charge is at the center of a square of side 10 cm. Find the work done in moving a charge of 10 µC between two diagonally opposite points on the square.

Solution:

Because these two locations are at equipotential, the work done in transporting a charge of 10 C between two diagonally opposite spots on the square will be zero.

Question 4: “For any charge configuration, equipotential surface through a point is normal to the electric field.” Justify.

Solution:

Work done in moving a charge over an equipotential surface is zero, hence a point on it will be normal to the electric field.

W = Fs cos θ 

∴ cos θ = 0 

or 

θ = 90^o

Question 5: Why must the electrostatic potential inside a hollow charged conductor be the same at every point?

Solution:

Because the electric field inside the hollow charged conductor is zero, no work is done in moving a small test charge within the conductor. As a result, the electrostatic potential inside a hollow charged conductor remains constant.