To find the energy of the system consisting of two charges \( q_1 = 4 \, \mu C \) and \( q_2 = -4 \, \mu C \) placed at coordinates \((-3, 0, 0)\) cm and \((3, 0, 0)\) cm respectively in an external electric field given by \( E = \frac{9 \times 10^6}{r^2} \, \text{C m}^{-2} \), we can follow these steps:
### Step 1: Calculate the potential due to the external electric field at the positions of the charges.
The potential \( V \) due to an electric field \( E \) is given by the formula:
\[
V = -\int E \, dr
\]
For the given electric field \( E = \frac{9 \times 10^6}{r^2} \), we can find the potential at a distance \( r \) from the origin.
### Step 2: Determine the potential at the position of \( q_1 \) and \( q_2 \).
The distance \( r_1 \) for charge \( q_1 \) (located at \((-3, 0, 0)\) cm) is \( 3 \, \text{cm} = 3 \times 10^{-2} \, \text{m} \).
The potential at \( q_1 \) is:
\[
V_1 = \frac{9 \times 10^6}{3 \times 10^{-2}} = 3 \times 10^8 \, \text{V}
\]
The distance \( r_2 \) for charge \( q_2 \) (located at \((3, 0, 0)\) cm) is also \( 3 \, \text{cm} = 3 \times 10^{-2} \, \text{m} \).
The potential at \( q_2 \) is:
\[
V_2 = \frac{9 \times 10^6}{3 \times 10^{-2}} = 3 \times 10^8 \, \text{V}
\]
### Step 3: Calculate the energy of each charge in the external field.
The energy \( U \) of a charge \( q \) in a potential \( V \) is given by:
\[
U = qV
\]
For charge \( q_1 \):
\[
U_1 = q_1 V_1 = (4 \times 10^{-6} \, \text{C})(3 \times 10^8 \, \text{V}) = 1.2 \, \text{J}
\]
For charge \( q_2 \):
\[
U_2 = q_2 V_2 = (-4 \times 10^{-6} \, \text{C})(3 \times 10^8 \, \text{V}) = -1.2 \, \text{J}
\]
### Step 4: Calculate the interaction energy between the two charges.
The interaction energy \( U_{int} \) between two point charges is given by:
\[
U_{int} = k \frac{q_1 q_2}{r}
\]
where \( k = \frac{1}{4\pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( r \) is the distance between the charges.
The distance \( r \) between \( q_1 \) and \( q_2 \) is:
\[
r = 6 \, \text{cm} = 6 \times 10^{-2} \, \text{m}
\]
Thus, the interaction energy is:
\[
U_{int} = k \frac{(4 \times 10^{-6})(-4 \times 10^{-6})}{6 \times 10^{-2}} = 9 \times 10^9 \frac{-16 \times 10^{-12}}{6 \times 10^{-2}} = -2.4 \, \text{J}
\]
### Step 5: Calculate the total energy of the system.
The total energy \( U_{total} \) of the system is the sum of the energies of the individual charges and the interaction energy:
\[
U_{total} = U_1 + U_2 + U_{int} = 1.2 \, \text{J} - 1.2 \, \text{J} - 2.4 \, \text{J} = -2.4 \, \text{J}
\]
### Final Answer:
The total energy of the system in the external field is:
\[
\boxed{-2.4 \, \text{J}}
\]
