Two point charges of `+10muC` and `+20muC` are placed in free space 2 cm apart. Find the electric potential at the middle point of the line joining the two charges.

Answer

To find the electric potential at the midpoint of the line joining two point charges of +10 µC and +20 µC placed 2 cm apart, we can follow these steps:

### Step 1: Identify the midpoint
The two charges are separated by a distance of 2 cm. Therefore, the midpoint (C) is located 1 cm (or 0.01 m) from each charge.

### Step 2: Use the formula for electric potential
The electric potential (V) due to a point charge is given by the formula:
\[ V = \frac{k \cdot Q}{r} \]
where:
- \( V \) is the electric potential,
- \( k \) is Coulomb's constant (\( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( Q \) is the charge,
- \( r \) is the distance from the charge to the point where the potential is being calculated.

### Step 3: Calculate the potential due to the +10 µC charge
For the +10 µC charge:
- \( Q_1 = 10 \times 10^{-6} \, \text{C} \)
- \( r_1 = 0.01 \, \text{m} \)

Plugging in the values:
\[ V_1 = \frac{9 \times 10^9 \cdot 10 \times 10^{-6}}{0.01} \]
\[ V_1 = \frac{9 \times 10^9 \cdot 10^{-5}}{0.01} \]
\[ V_1 = 9 \times 10^4 \, \text{V} \]

### Step 4: Calculate the potential due to the +20 µC charge
For the +20 µC charge:
- \( Q_2 = 20 \times 10^{-6} \, \text{C} \)
- \( r_2 = 0.01 \, \text{m} \)

Now, calculate \( V_2 \):
\[ V_2 = \frac{9 \times 10^9 \cdot 20 \times 10^{-6}}{0.01} \]
\[ V_2 = \frac{9 \times 10^9 \cdot 2 \times 10^{-5}}{0.01} \]
\[ V_2 = 18 \times 10^4 \, \text{V} \]

### Step 5: Find the total electric potential at the midpoint
The total electric potential \( V \) at point C is the sum of the potentials due to both charges:
\[ V = V_1 + V_2 \]
\[ V = 9 \times 10^4 + 18 \times 10^4 \]
\[ V = 27 \times 10^4 \, \text{V} \]

### Final Answer
The electric potential at the midpoint of the line joining the two charges is:
\[ V = 27 \times 10^4 \, \text{V} \]
or
\[ V = 270000 \, \text{V} \]

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