Two point particle A and B having charges of `4xx10^(-6) C` and `-64xx10^(-6) C` respectively are held at a separation of 90 cm. Locate the point(s) on the line AB or on its extension where the electric field is zero

Answer

To find the point(s) on the line AB or its extension where the electric field is zero due to the charges A and B, we can follow these steps:

### Step 1: Understand the Configuration
We have two point charges:
- Charge A: \( q_A = 4 \times 10^{-6} \, C \) (positive)
- Charge B: \( q_B = -64 \times 10^{-6} \, C \) (negative)
- Distance between A and B: \( d = 90 \, cm = 0.9 \, m \)

### Step 2: Set Up the Electric Field Equation
We need to find a point C on the line AB or its extension where the electric field is zero. Let’s denote the distance from charge A to point C as \( x \). Therefore, the distance from charge B to point C will be \( 0.9 + x \) if C is between A and B, or \( |x - 0.9| \) if C is outside the segment AB.

The electric field \( E \) due to a point charge is given by:
\[ E = k \frac{|q|}{r^2} \]
where \( k \) is Coulomb's constant.

### Step 3: Electric Field Contributions
1. **If C is between A and B (0 < x < 0.9)**:
- Electric field due to A at C:
\[ E_A = k \frac{4 \times 10^{-6}}{x^2} \]
- Electric field due to B at C:
\[ E_B = k \frac{64 \times 10^{-6}}{(0.9 - x)^2} \]
- Since B is negative, its electric field direction is towards B (opposite to A):
\[ E_B = -k \frac{64 \times 10^{-6}}{(0.9 - x)^2} \]

Setting the total electric field to zero:
\[ E_A + E_B = 0 \]
\[ k \frac{4 \times 10^{-6}}{x^2} - k \frac{64 \times 10^{-6}}{(0.9 - x)^2} = 0 \]

### Step 4: Simplifying the Equation
Cancel \( k \) from both sides:
\[ \frac{4 \times 10^{-6}}{x^2} = \frac{64 \times 10^{-6}}{(0.9 - x)^2} \]

Cross-multiplying gives:
\[ 4 \times (0.9 - x)^2 = 64 \times x^2 \]

### Step 5: Expanding and Rearranging
Expanding the left side:
\[ 4(0.81 - 1.8x + x^2) = 64x^2 \]
\[ 3.24 - 7.2x + 4x^2 = 64x^2 \]

Rearranging gives:
\[ 60x^2 + 7.2x - 3.24 = 0 \]

### Step 6: Solving the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here \( a = 60, b = 7.2, c = -3.24 \)

Calculating the discriminant:
\[ b^2 - 4ac = (7.2)^2 - 4 \cdot 60 \cdot (-3.24) \]
\[ = 51.84 + 777.6 = 829.44 \]

Now substituting into the quadratic formula:
\[ x = \frac{-7.2 \pm \sqrt{829.44}}{120} \]
\[ x = \frac{-7.2 \pm 28.8}{120} \]

Calculating the two possible values for x:
1. \( x_1 = \frac{21.6}{120} = 0.18 \, m \) (valid, between A and B)
2. \( x_2 = \frac{-36}{120} = -0.3 \, m \) (not valid, outside the segment)

### Step 7: Conclusion
The point where the electric field is zero is at a distance of \( 0.18 \, m \) from charge A towards charge B.