Two point charges `4mu C and 9muC` are separated by 50cm. 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)V`
C. `9 xx 10^(4)V`
D. zero

Explanation

To solve the problem step by step, we will find the point between the two charges where the electric field is zero and then calculate the electric potential at that point.

### Step 1: Understand the Configuration
We have two point charges:
- Charge \( Q_1 = 4 \, \mu C = 4 \times 10^{-6} \, C \)
- Charge \( Q_2 = 9 \, \mu C = 9 \times 10^{-6} \, C \)

These charges are separated by a distance of \( d = 50 \, cm = 0.5 \, m \).

### Step 2: Determine the Point Where Electric Field is Zero
The electric field due to a point charge is given by:
\[ E = \frac{k \cdot |Q|}{r^2} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \).

To find the point where the electric field is zero, we need to set the electric fields due to both charges equal to each other. Let's denote the distance from \( Q_1 \) to the point where the electric field is zero as \( x \). Therefore, the distance from \( Q_2 \) to that point will be \( (0.5 - x) \).

Setting the magnitudes of the electric fields equal:
\[ \frac{k \cdot 4 \times 10^{-6}}{x^2} = \frac{k \cdot 9 \times 10^{-6}}{(0.5 - x)^2} \]

### Step 3: Simplify the Equation
We can cancel \( k \) from both sides:
\[ \frac{4 \times 10^{-6}}{x^2} = \frac{9 \times 10^{-6}}{(0.5 - x)^2} \]

Cross-multiplying gives:
\[ 4 \times 10^{-6} \cdot (0.5 - x)^2 = 9 \times 10^{-6} \cdot x^2 \]

Dividing both sides by \( 10^{-6} \):
\[ 4(0.5 - x)^2 = 9x^2 \]

### Step 4: Expand and Rearrange
Expanding the left side:
\[ 4(0.25 - x + x^2) = 9x^2 \]
\[ 1 - 4x + 4x^2 = 9x^2 \]

Rearranging gives:
\[ 0 = 5x^2 + 4x - 1 \]

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 5, b = 4, c = -1 \):
\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 5 \cdot (-1)}}{2 \cdot 5} \]
\[ x = \frac{-4 \pm \sqrt{16 + 20}}{10} \]
\[ x = \frac{-4 \pm \sqrt{36}}{10} \]
\[ x = \frac{-4 \pm 6}{10} \]

Calculating the two possible values:
1. \( x = \frac{2}{10} = 0.2 \, m = 20 \, cm \)
2. \( x = \frac{-10}{10} = -1 \, m \) (not physically meaningful)

Thus, the point where the electric field is zero is \( 20 \, cm \) from the \( 4 \, \mu C \) charge.

### Step 6: Calculate the Electric Potential at that Point
The electric potential \( V \) at a point due to a point charge is given by:
\[ V = k \cdot \frac{Q}{r} \]

The total potential at the point \( 20 \, cm \) from \( Q_1 \) (and \( 30 \, cm \) from \( Q_2 \)):
\[ V = V_1 + V_2 = k \cdot \frac{Q_1}{0.2} + k \cdot \frac{Q_2}{0.3} \]
Substituting values:
\[ V = \frac{1}{4 \pi \epsilon_0} \left( \frac{4 \times 10^{-6}}{0.2} + \frac{9 \times 10^{-6}}{0.3} \right) \]

Calculating:
\[ V = \frac{9 \times 10^9}{4 \pi} \left( 20 \times 10^{-6} + 30 \times 10^{-6} \right) \]
\[ V = \frac{9 \times 10^9}{4 \pi} \cdot 50 \times 10^{-6} \]

### Final Calculation
Calculating the numerical value will give the potential at that point.