Q.(a) Determine the electrostatic potential energy of a system consisting of two charges $7\ \mu\text{C}$ and $-2\ \mu\text{C}$ (and with no external field) placed at $(-9\ \text{cm}, 0, 0)$ and $(9\ \text{cm}, 0, 0)$ respectively.

Concept: Electric Potential Energy — the work done to assemble a system of charges from infinity, or the energy stored in the configuration.

(a) For two point charges, the electrostatic potential energy is

Here $q_1 = 7\ \mu\text{C}$, $q_2 = -2\ \mu\text{C}$, and separation $r = 18\ \text{cm} = 0.18\ \text{m}$.

Using $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{Nm}^2/\text{C}^2$:

$U = 9 \times 10^9 \times \frac{(7 \times 10^{-6})(-2 \times 10^{-6})}{0.18} = 9 \times 10^9 \times \frac{-14 \times 10^{-12}}{0.18}$ …

Let’s unpack this step by step. The core idea is that electrostatic potential energy is the work done by an external agent to bring charges from infinite separation to their final positions, against the electric forces between them. When there’s no external field, only the mutual interaction matters. When an external field is present, we must also account for the work done to place each charge into that field.

(a) Potential energy of the two-charge system

1. Identify the charges and separation.

We have $q_1 = 7\ \mu\text{C} = 7 \times 10^{-6}\ \text{C}$ and $q_2 = -2\ \mu\text{C} = -2 \times 10^{-6}\ \text{C}$. Their positions are $(-9\ \text{cm}, 0, 0)$ and $(9\ \text{cm}, 0, 0)$, so the distance between them is $r = 18\ \text{cm} = 0.18\ \text{m}$.

2. Recall the formula for potential energy of a pair of point charges.

For two point charges in free space, with no external field, the electrostatic potential energy is:

Here $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{N m}^2/\text{C}^2$.

3. Plug in the numbers.

First compute the numerator: $7 \times (-2) = -14$, and $10^{-6} \times 10^{-6} = 10^{-12}$, so $q_1 q_2 = -14 \times 10^{-12}\ \text{C}^2$.

Now:

Simplify $\frac{-14}{0.18} = -\frac{1400}{18} = -\frac{700}{9} \approx -77.78$.

So:

Result for (a): $U = -0.7\ \text{J}$.

(b) Work required to separate them infinitely

4. Understand what “separate infinitely” means.

When the charges are infinitely far apart, their mutual potential energy becomes zero (since $r \to \infty$). The work done by an external agent to move them from their initial configuration to infinity equals the change in potential energy:

The positive sign means work must be done on the system (against the attractive force) to pull them apart.

Result for (b): $W = 0.7\ \text{J}$.

(c) Energy in an external electric field

5. The new situation.

Now the same two charges are placed in an external field $\mathbf{E} = A \frac{1}{r^2} \hat{r}$, with $A = 9 \times 10^{5}\ \text{N C}^{-1} \text{m}^2$. This is a radial field — like that of a point charge at the origin, but here it’s externally imposed. The potential corresponding to this field is:

We can take the constant as zero at infinity, so $V_{\text{ext}}(r) = \frac{A}{r}$.

6. Total electrostatic energy of the configuration.

When an external field is present, the total electrostatic energy has two contributions:

• The self-energy of the charges in the external field: for each charge $q$, this is $q \, V_{\text{ext}}(r)$. …

Method: Total Electrostatic Energy of a Charge System Placed in an External Field

This method applies whenever a system of point charges has both (i) its own mutual interaction energy, and (ii) an externally-imposed field acting on it, and you're asked for the combined energy.

Steps

Step 1: Compute the mutual (interaction) energy of the charges alone

Treat the given charges as an isolated system first, ignoring the external field, and sum the pairwise potential energy:

Step 2: Find the external potential $V_{\text{ext}}$ at each charge's location

If the external field $\vec{E}$ is given as a function of position, recover its potential by integrating:

taking the reference $V_{\text{ext}} = 0$ at infinity unless told otherwise.

Step 3: Compute each charge's energy of interaction with the external field …