Two tiny spheres carrying charges `1.8 muC and 2.8 mu C` are located at 40 cm apart. The potential at the mid-point of the line joining the two charges is
In the above question, the potential at a point20cm from the mid-point of the line joining the two charges in a place normal to the line and passing through the mid-point is

Options

A. `1.4 xx 10^(5)V`
B. `4.2 xx 10^(3)V`
C. `2.9 xx 10^(4)V`
D. `3.7 xx 10^(5)V`

Explanation

To solve the problem, we need to calculate the electric potential at a point located 20 cm from the midpoint of the line joining two charged spheres, each carrying charges of \(1.8 \, \mu C\) and \(2.8 \, \mu C\), which are 40 cm apart.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- Let \(Q_1 = 1.8 \, \mu C = 1.8 \times 10^{-6} \, C\)
- Let \(Q_2 = 2.8 \, \mu C = 2.8 \times 10^{-6} \, C\)
- The distance between the charges is \(d = 40 \, cm = 0.4 \, m\).

2. **Determine the Midpoint:**
- The midpoint between the two charges is at a distance of \(20 \, cm = 0.2 \, m\) from each charge.

3. **Calculate the Distance to the Point of Interest:**
- The point of interest is located \(20 \, cm\) from the midpoint, in a direction perpendicular to the line joining the charges.
- Using the Pythagorean theorem, the distance \(r\) from each charge to the point of interest can be calculated as:
\[ r = \sqrt{(0.2)^2 + (0.2)^2} = \sqrt{0.04 + 0.04} = \sqrt{0.08} = 0.2828 \, m \]

4. **Calculate the Electric Potential at the Point:**
- The electric potential \(V\) at a point due to a point charge is given by:
\[ V = k \cdot \frac{Q}{r} \]
- Where \(k = 9 \times 10^9 \, N \cdot m^2/C^2\).
- The total potential at the point due to both charges is:
\[ V_P = V_1 + V_2 = k \cdot \frac{Q_1}{r} + k \cdot \frac{Q_2}{r} \]
- Since \(r\) is the same for both charges:
\[ V_P = k \left( \frac{Q_1 + Q_2}{r} \right) \]

5. **Substituting the Values:**
- First, calculate \(Q_1 + Q_2\):
\[ Q_1 + Q_2 = 1.8 \times 10^{-6} + 2.8 \times 10^{-6} = 4.6 \times 10^{-6} \, C \]
- Now substitute the values into the potential formula:
\[ V_P = 9 \times 10^9 \cdot \frac{4.6 \times 10^{-6}}{0.2828} \]
- Calculate \(V_P\):
\[ V_P \approx 9 \times 10^9 \cdot 1.627 \times 10^{-5} \approx 146.3 \times 10^3 \, V \approx 1.463 \times 10^5 \, V \]

### Final Answer:
The potential at the point 20 cm from the midpoint of the line joining the two charges is approximately \(1.463 \times 10^5 \, V\).