two point charges + 16 q and -4q located at x=0 and x=L respectively. the location of a point on the x-axis from x=0, at which the net electric field due to these two charges is zero is

Options

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

Answer

To find the location on the x-axis where the net electric field due to the two point charges \( +16q \) at \( x = 0 \) and \( -4q \) at \( x = L \) is zero, we can follow these steps:

### Step 1: Understand the Electric Field Directions
The electric field due to a positive charge points away from the charge, while the electric field due to a negative charge points towards the charge. Therefore:
- The electric field \( E_1 \) due to \( +16q \) at a point \( x \) on the x-axis will point to the right (away from the charge).
- The electric field \( E_2 \) due to \( -4q \) will also point to the right (towards the charge).

### Step 2: Set Up the Electric Field Equations
The electric field \( E \) due to a point charge is given by the formula:
\[ E = \frac{k \cdot |q|}{r^2} \]
where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge to the point where the electric field is being calculated.

For the charge \( +16q \) at \( x = 0 \):
\[ E_1 = \frac{k \cdot 16q}{x^2} \]

For the charge \( -4q \) at \( x = L \):
\[ E_2 = \frac{k \cdot 4q}{(L - x)^2} \]

### Step 3: Set the Electric Fields Equal to Each Other
To find the point where the net electric field is zero, we set \( E_1 = E_2 \):
\[ \frac{k \cdot 16q}{x^2} = \frac{k \cdot 4q}{(L - x)^2} \]

### Step 4: Simplify the Equation
We can cancel \( k \) and \( q \) from both sides:
\[ \frac{16}{x^2} = \frac{4}{(L - x)^2} \]

Cross-multiplying gives:
\[ 16(L - x)^2 = 4x^2 \]

### Step 5: Expand and Rearrange
Expanding the left side:
\[ 16(L^2 - 2Lx + x^2) = 4x^2 \]
This simplifies to:
\[ 16L^2 - 32Lx + 16x^2 = 4x^2 \]
Rearranging gives:
\[ 12x^2 - 32Lx + 16L^2 = 0 \]

### Step 6: Solve the Quadratic Equation
We can simplify this quadratic equation:
\[ 3x^2 - 8Lx + 4L^2 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 3 \), \( b = -8L \), and \( c = 4L^2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-8L)^2 - 4 \cdot 3 \cdot 4L^2 = 64L^2 - 48L^2 = 16L^2 \]

Now substituting into the quadratic formula:
\[ x = \frac{8L \pm 4L}{6} \]

This gives two solutions:
1. \( x = \frac{12L}{6} = 2L \)
2. \( x = \frac{4L}{6} = \frac{2L}{3} \)

### Step 7: Determine Valid Solutions
The point \( x = 2L \) is outside the region between the two charges, while \( x = \frac{2L}{3} \) lies between them. Hence, the location on the x-axis where the net electric field is zero is:
\[ x = \frac{2L}{3} \]

### Final Answer
The location of the point on the x-axis where the net electric field is zero is:
\[ \boxed{\frac{2L}{3}} \]