Two point charges `+8q` and `-2q` are located at `x=0` and `x=L` respectively. The location of a point on the x axis at which the net electric field due to these two point charges is zero is

Options

A. (a) `L/4`
B. (b) `2L`
C. (c) `4L`
D. (d) `8L`

Explanation

To find the location on the x-axis where the net electric field due to the two point charges \( +8q \) and \( -2q \) is zero, we can follow these steps:

### Step 1: Identify the positions of the charges
- The charge \( +8q \) is located at \( x = 0 \).
- The charge \( -2q \) is located at \( x = L \).

### Step 2: Set up the electric field equations
Let \( P \) be a point on the x-axis where the electric field is zero. We will consider two cases for the location of point \( P \):
1. To the left of charge \( +8q \) (i.e., \( x < 0 \))
2. Between the two charges (i.e., \( 0 < x < L \))
3. To the right of charge \( -2q \) (i.e., \( x > L \))

### Step 3: Analyze the case where \( P \) is between the charges
Assuming \( P \) is between the charges, we can denote its position as \( x \). The electric field \( E_1 \) due to the charge \( +8q \) at point \( P \) is given by:
\[ E_1 = \frac{k \cdot 8q}{x^2} \]
The electric field \( E_2 \) due to the charge \( -2q \) at point \( P \) is given by:
\[ E_2 = \frac{k \cdot 2q}{(L - x)^2} \]
Since \( E_1 \) points away from \( +8q \) and \( E_2 \) points towards \( -2q \), we can set up the equation:
\[ E_1 = E_2 \]
This leads to:
\[ \frac{k \cdot 8q}{x^2} = \frac{k \cdot 2q}{(L - x)^2} \]

### Step 4: Simplify the equation
Cancel \( k \) and \( q \) from both sides:
\[ \frac{8}{x^2} = \frac{2}{(L - x)^2} \]
Cross-multiplying gives:
\[ 8(L - x)^2 = 2x^2 \]
Dividing both sides by 2:
\[ 4(L - x)^2 = x^2 \]

### Step 5: Expand and rearrange the equation
Expanding the left side:
\[ 4(L^2 - 2Lx + x^2) = x^2 \]
This simplifies to:
\[ 4L^2 - 8Lx + 4x^2 = x^2 \]
Rearranging gives:
\[ 3x^2 - 8Lx + 4L^2 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 3 \), \( b = -8L \), and \( c = 4L^2 \):
\[ x = \frac{8L \pm \sqrt{(-8L)^2 - 4 \cdot 3 \cdot 4L^2}}{2 \cdot 3} \]
\[ x = \frac{8L \pm \sqrt{64L^2 - 48L^2}}{6} \]
\[ x = \frac{8L \pm \sqrt{16L^2}}{6} \]
\[ x = \frac{8L \pm 4L}{6} \]
This gives two possible solutions:
1. \( x = \frac{12L}{6} = 2L \)
2. \( x = \frac{4L}{6} = \frac{2L}{3} \)

### Step 7: Determine the valid solution
Since \( x \) must be between \( 0 \) and \( L \) for the case we considered, the valid solution is:
\[ x = \frac{2L}{3} \]

### Conclusion
The location on the x-axis where the net electric field due to the two point charges is zero is:
\[ \boxed{\frac{2L}{3}} \]