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

Options

A. `3.8xx10^(4)V`
B. `2.1xx10^(5)V`
C. `4.3xx10^(4)V`
D. `6.3xx10^(5)V`

Explanation

To find the electric potential at the midpoint of the line joining two charges, we can follow these steps:

### Step 1: Understand the Problem
We have two charges:
- Charge \( Q_1 = 1.8 \, \mu C = 1.8 \times 10^{-6} \, C \)
- Charge \( Q_2 = 2.8 \, \mu C = 2.8 \times 10^{-6} \, C \)

The distance between the two charges is \( d = 40 \, cm = 0.4 \, m \).

### Step 2: Find the Midpoint
The midpoint \( O \) is located at a distance of \( \frac{d}{2} = \frac{40 \, cm}{2} = 20 \, cm = 0.2 \, m \) from each charge.

### Step 3: Use the Formula for Electric Potential
The electric potential \( V \) at a point due to a point charge is given by:
\[ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{r} \]
where:
- \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \)
- \( r \) is the distance from the charge to the point where the potential is being calculated.

### Step 4: Calculate the Potential at the Midpoint
The total potential \( V_O \) at point \( O \) due to both charges is:
\[ V_O = V_1 + V_2 \]
where:
- \( V_1 \) is the potential due to charge \( Q_1 \)
- \( V_2 \) is the potential due to charge \( Q_2 \)

Calculating \( V_1 \):
\[ V_1 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_1}{R_1} = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{1.8 \times 10^{-6}}{0.2} \]

Calculating \( V_2 \):
\[ V_2 = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_2}{R_2} = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{2.8 \times 10^{-6}}{0.2} \]

### Step 5: Combine the Potentials
Now, combine \( V_1 \) and \( V_2 \):
\[ V_O = V_1 + V_2 \]

### Step 6: Substitute Values and Calculate
Substituting the values into the equations:
\[ V_O = \left( \frac{9 \times 10^9}{0.2} \cdot (1.8 \times 10^{-6}) \right) + \left( \frac{9 \times 10^9}{0.2} \cdot (2.8 \times 10^{-6}) \right) \]
\[ = 9 \times 10^9 \left( \frac{1.8 \times 10^{-6}}{0.2} + \frac{2.8 \times 10^{-6}}{0.2} \right) \]
\[ = 9 \times 10^9 \left( \frac{1.8 + 2.8}{0.2} \times 10^{-6} \right) \]
\[ = 9 \times 10^9 \left( \frac{4.6 \times 10^{-6}}{0.2} \right) \]
\[ = 9 \times 10^9 \times 23 \times 10^{-6} \]
\[ = 2070 \times 10^3 \, V \]
\[ = 2.07 \times 10^6 \, V \]

### Step 7: Final Answer
Thus, the potential at the midpoint \( O \) is approximately \( 2.1 \times 10^5 \, V \).