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

Options

A. 2L
B. L/4
C. 8L
D. 4L

Answer

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 equation for the electric field
- Let \( x \) be the distance from the charge \( +8q \) where the electric field is zero.
- The distance from the charge \( -2q \) to this point will then be \( (x - L) \).

### Step 3: Write the expression for the electric fields
- The electric field \( E_1 \) due to the charge \( +8q \) at a distance \( x \) is given by:
\[ E_1 = \frac{k \cdot 8q}{x^2} \]
- The electric field \( E_2 \) due to the charge \( -2q \) at a distance \( (x - L) \) is given by:
\[ E_2 = \frac{k \cdot 2q}{(x - L)^2} \]

### Step 4: Set the magnitudes of the electric fields equal
Since the net electric field is zero, we set the magnitudes of \( E_1 \) and \( E_2 \) equal to each other:
\[ \frac{k \cdot 8q}{x^2} = \frac{k \cdot 2q}{(x - L)^2} \]

### Step 5: Simplify the equation
We can cancel \( k \) and \( q \) from both sides (assuming \( q \neq 0 \)):
\[ \frac{8}{x^2} = \frac{2}{(x - L)^2} \]

### Step 6: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 8(x - L)^2 = 2x^2 \]

### Step 7: Expand and rearrange the equation
Expanding the left side:
\[ 8(x^2 - 2xL + L^2) = 2x^2 \]
This simplifies to:
\[ 8x^2 - 16xL + 8L^2 = 2x^2 \]
Rearranging gives:
\[ 6x^2 - 16xL + 8L^2 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 6 \), \( b = -16L \), and \( c = 8L^2 \):
\[ x = \frac{16L \pm \sqrt{(-16L)^2 - 4 \cdot 6 \cdot 8L^2}}{2 \cdot 6} \]
Calculating the discriminant:
\[ x = \frac{16L \pm \sqrt{256L^2 - 192L^2}}{12} \]
\[ x = \frac{16L \pm \sqrt{64L^2}}{12} \]
\[ x = \frac{16L \pm 8L}{12} \]

### Step 9: Find the two possible values for \( x \)
Calculating the two possible values:
1. \( x = \frac{24L}{12} = 2L \)
2. \( x = \frac{8L}{12} = \frac{2L}{3} \)

### Step 10: Determine the valid solution
Since the point must be to the left of \( -2q \) (which is at \( x = L \)), the valid solution is:
\[ x = \frac{2L}{3} \]

### Final Answer
The location of the point on the x-axis from \( +8q \) at which the net electric field is zero is:
\[ x = \frac{2L}{3} \]