Two point charges 4`mu`C and 9`mu`C are separated by 50 cm. The potential at the point between them where the field has zero strength is

Options

A. `4.5 xx 10^5 v`
B. `9 xx 10^5 K`
C. `9 xx 10^4 V`
D. Zero

Answer

To solve the problem, we need to find the potential at the point between two charges where the electric field is zero. We have two point charges: \( q_1 = 4 \, \mu C \) and \( q_2 = 9 \, \mu C \) separated by a distance of \( 50 \, cm \).

### Step-by-Step Solution:

1. **Identify the Charges and Distance**:
- Let \( q_1 = 4 \, \mu C = 4 \times 10^{-6} \, C \)
- Let \( q_2 = 9 \, \mu C = 9 \times 10^{-6} \, C \)
- Distance between the charges, \( d = 50 \, cm = 0.5 \, m \)

2. **Set Up the Equation for Electric Field**:
- Let the point where the electric field is zero be at a distance \( x \) from \( q_1 \) (4 µC).
- The distance from \( q_2 \) (9 µC) will then be \( 50 - x \) cm.
- The electric field due to \( q_1 \) at this point is given by:
\[ E_1 = \frac{k \cdot q_1}{x^2} \]
- The electric field due to \( q_2 \) at this point is given by:
\[ E_2 = \frac{k \cdot q_2}{(50 - x)^2} \]
- For the electric field to be zero, these two fields must be equal:
\[ \frac{k \cdot q_1}{x^2} = \frac{k \cdot q_2}{(50 - x)^2} \]
- We can cancel \( k \) from both sides:
\[ \frac{q_1}{x^2} = \frac{q_2}{(50 - x)^2} \]

3. **Substitute the Values**:
- Substitute \( q_1 \) and \( q_2 \):
\[ \frac{4 \times 10^{-6}}{x^2} = \frac{9 \times 10^{-6}}{(50 - x)^2} \]

4. **Cross Multiply**:
- Cross multiplying gives:
\[ 4 \times 10^{-6} \cdot (50 - x)^2 = 9 \times 10^{-6} \cdot x^2 \]
- Dividing by \( 10^{-6} \):
\[ 4(50 - x)^2 = 9x^2 \]

5. **Expand and Rearrange**:
- Expanding the left side:
\[ 4(2500 - 100x + x^2) = 9x^2 \]
\[ 10000 - 400x + 4x^2 = 9x^2 \]
- Rearranging gives:
\[ 5x^2 + 400x - 10000 = 0 \]

6. **Solve the Quadratic Equation**:
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 5, b = -400, c = -10000 \)
- Calculate the discriminant:
\[ D = (-400)^2 - 4 \cdot 5 \cdot (-10000) = 160000 + 200000 = 360000 \]
- Now, substituting into the formula:
\[ x = \frac{400 \pm \sqrt{360000}}{10} = \frac{400 \pm 600}{10} \]
- This gives two solutions:
\[ x = 100 \, cm \quad \text{(not valid, as it exceeds 50 cm)} \]
\[ x = -20 \, cm \quad \text{(not valid)} \]
- Hence, we check the valid position:
\[ x = 20 \, cm \quad \text{(valid)} \]

7. **Calculate the Potential at the Point**:
- The potential \( V \) at the point where the electric field is zero is given by:
\[ V = V_1 + V_2 \]
- Where:
\[ V_1 = \frac{k \cdot q_1}{x} = \frac{9 \times 10^9 \cdot 4 \times 10^{-6}}{0.2} \]
\[ V_2 = \frac{k \cdot q_2}{(50 - x)} = \frac{9 \times 10^9 \cdot 9 \times 10^{-6}}{0.3} \]

8. **Final Calculation**:
- Calculate \( V_1 \) and \( V_2 \):
\[ V_1 = \frac{9 \times 10^9 \cdot 4 \times 10^{-6}}{0.2} = 180000 \, V \]
\[ V_2 = \frac{9 \times 10^9 \cdot 9 \times 10^{-6}}{0.3} = 270000 \, V \]
- Therefore:
\[ V = V_1 + V_2 = 180000 + 270000 = 450000 \, V = 4.5 \times 10^5 \, V \]

### Final Answer:
The potential at the point between the charges where the electric field is zero is \( 4.5 \times 10^5 \, V \).