Three point charge `q_1=1muC, q_2=-2muC` and `q_3=3muC` are placed at `(1m, 0,0), (0,2m,0)` and `(0,0,3m)` respectively. Find the electric potential at the origin.

Options

A. `2.0xx10^3V`
B. `9.0xx10^3V`
C. `1.0xx10^3V`
D. `10xx10^3V`

Explanation

To find the electric potential at the origin due to the three point charges \( q_1, q_2, \) and \( q_3 \), we can follow these steps:

### Step 1: Identify the charges and their positions
- \( q_1 = 1 \mu C \) is located at \( (1m, 0, 0) \)
- \( q_2 = -2 \mu C \) is located at \( (0, 2m, 0) \)
- \( q_3 = 3 \mu C \) is located at \( (0, 0, 3m) \)

### Step 2: Calculate the distances from the origin to each charge
- The distance from the origin to \( q_1 \) is:
\[ R_1 = 1 \text{ m} \]
- The distance from the origin to \( q_2 \) is:
\[ R_2 = 2 \text{ m} \]
- The distance from the origin to \( q_3 \) is:
\[ R_3 = 3 \text{ m} \]

### Step 3: Use the formula for electric potential
The electric potential \( V \) at a point due to a point charge is given by:
\[ V = k \frac{Q}{R} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \text{ N m}^2/\text{C}^2 \).

### Step 4: Calculate the total electric potential at the origin
The total electric potential at the origin due to all three charges is the sum of the potentials due to each charge:
\[ V = V_1 + V_2 + V_3 \]
Substituting the values:
\[ V_1 = k \frac{q_1}{R_1} = 9 \times 10^9 \cdot \frac{1 \times 10^{-6}}{1} = 9 \times 10^3 \text{ V} \]
\[ V_2 = k \frac{q_2}{R_2} = 9 \times 10^9 \cdot \frac{-2 \times 10^{-6}}{2} = -9 \times 10^3 \text{ V} \]
\[ V_3 = k \frac{q_3}{R_3} = 9 \times 10^9 \cdot \frac{3 \times 10^{-6}}{3} = 9 \times 10^3 \text{ V} \]

### Step 5: Add the potentials together
Now we can sum these potentials:
\[ V = V_1 + V_2 + V_3 = 9 \times 10^3 - 9 \times 10^3 + 9 \times 10^3 = 9 \times 10^3 \text{ V} \]

### Final Answer
The electric potential at the origin is:
\[ \boxed{9.0 \times 10^3 \text{ V}} \]