Two points charges are located on the x-axis, `q_1=-1muC` at `x=0 and q_2=+1muc` at `x=1m`.
a. Find the work that must be done by an external force to bring a third point charge `q_3=+1muC` from infinity to `x=2m`.
b. Find the total potential energy of the system of three charges.

Explanation

To solve the problem step by step, we will break it down into two parts as indicated in the question.

### Part (a): Work Done by External Force

1. **Identify the Charges and Their Positions:**
- Charge \( q_1 = -1 \, \mu C \) at \( x = 0 \, m \)
- Charge \( q_2 = +1 \, \mu C \) at \( x = 1 \, m \)
- Charge \( q_3 = +1 \, \mu C \) is to be brought from infinity to \( x = 2 \, m \)

2. **Calculate the Final Potential Energy (U_final):**
- The potential energy of the system when \( q_3 \) is at \( x = 2 \, m \) can be calculated using the formula for potential energy between point charges:
\[ U_{final} = k \frac{q_1 q_2}{r_{12}} + k \frac{q_2 q_3}{r_{23}} + k \frac{q_1 q_3}{r_{13}} \]
- Where:
- \( r_{12} = 1 \, m \) (distance between \( q_1 \) and \( q_2 \))
- \( r_{23} = 1 \, m \) (distance between \( q_2 \) and \( q_3 \))
- \( r_{13} = 2 \, m \) (distance between \( q_1 \) and \( q_3 \))
- Substituting the values:
\[ U_{final} = k \frac{(-1 \times 10^{-6})(1 \times 10^{-6})}{1} + k \frac{(1 \times 10^{-6})(1 \times 10^{-6})}{1} + k \frac{(-1 \times 10^{-6})(1 \times 10^{-6})}{2} \]

3. **Calculate Each Term:**
- \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \)
- First term:
\[ k \frac{-1 \times 10^{-12}}{1} = -9 \times 10^{-3} \, J \]
- Second term:
\[ k \frac{1 \times 10^{-12}}{1} = 9 \times 10^{-3} \, J \]
- Third term:
\[ k \frac{-1 \times 10^{-12}}{2} = -4.5 \times 10^{-3} \, J \]

4. **Combine the Terms:**
\[ U_{final} = -9 \times 10^{-3} + 9 \times 10^{-3} - 4.5 \times 10^{-3} = -4.5 \times 10^{-3} \, J \]

5. **Calculate the Initial Potential Energy (U_initial):**
- When \( q_3 \) is at infinity, there is no interaction with \( q_1 \) and \( q_2 \), hence:
\[ U_{initial} = k \frac{q_1 q_2}{r_{12}} = k \frac{(-1 \times 10^{-6})(1 \times 10^{-6})}{1} = -9 \times 10^{-3} \, J \]

6. **Calculate the Work Done (W):**
- The work done by the external force is given by:
\[ W = U_{final} - U_{initial} \]
- Substituting the values:
\[ W = (-4.5 \times 10^{-3}) - (-9 \times 10^{-3}) = 4.5 \times 10^{-3} \, J \]

### Part (b): Total Potential Energy of the System

1. **Total Potential Energy (U_total):**
- The total potential energy of the system is simply the final potential energy calculated above:
\[ U_{total} = U_{final} = -4.5 \times 10^{-3} \, J \]

### Final Answers:
- (a) The work done by the external force is \( 4.5 \, mJ \) (or \( 4.5 \times 10^{-3} \, J \)).
- (b) The total potential energy of the system is \( -4.5 \, mJ \) (or \( -4.5 \times 10^{-3} \, J \)).