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 solve the problem of finding the location on the x-axis where the net electric field due to two point charges \( +16q \) and \( -4q \) is zero, we will follow these steps:

### Step 1: Understand the Configuration
We have two charges:
- Charge \( +16q \) located at \( x = 0 \)
- Charge \( -4q \) located at \( x = L \)

### Step 2: Identify the Regions
The electric field due to these charges will behave differently in various regions:
- **Region 1**: To the left of \( +16q \) (i.e., \( x < 0 \))
- **Region 2**: Between the two charges (i.e., \( 0 < x < L \))
- **Region 3**: To the right of \( -4q \) (i.e., \( x > L \))

### Step 3: Analyze the Electric Field Directions
- In **Region 1**, the electric field due to \( +16q \) points to the right (away from the charge), and the electric field due to \( -4q \) also points to the right (towards the charge). Therefore, the net electric field cannot be zero here.

- In **Region 2**, the electric field due to \( +16q \) points to the right, while the electric field due to \( -4q \) points to the left. Since both fields are in opposite directions, it is possible for them to cancel each other out.

- In **Region 3**, the electric field due to \( +16q \) points to the right, and the electric field due to \( -4q \) also points to the right (away from the negative charge). Thus, the net electric field cannot be zero here either.

### Step 4: Set Up the Equation for Electric Fields
We will find the point \( x \) in Region 2 where the electric fields due to both charges are equal in magnitude.

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

- The distance from \( +16q \) to point \( x \) is \( x \).
- The distance from \( -4q \) to point \( x \) is \( L - x \).

Setting the magnitudes of the electric fields equal gives:
\[ \frac{k \cdot 16q}{x^2} = \frac{k \cdot 4q}{(L - x)^2} \]

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

### Step 7: Solve the Quadratic Equation
This is a standard quadratic equation in the form \( ax^2 + bx + c = 0 \):
\[ 12x^2 - 32Lx + 16L^2 = 0 \]

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 12 \), \( b = -32L \), and \( c = 16L^2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-32L)^2 - 4 \cdot 12 \cdot 16L^2 = 1024L^2 - 768L^2 = 256L^2 \]

Now substituting back into the quadratic formula:
\[ x = \frac{32L \pm \sqrt{256L^2}}{2 \cdot 12} = \frac{32L \pm 16L}{24} \]

This gives us two possible solutions:
1. \( x = \frac{48L}{24} = 2L \)
2. \( x = \frac{16L}{24} = \frac{2L}{3} \)

### Step 8: Determine Valid Solution
Since we are looking for a point where the electric field is zero, we need to check the regions:
- \( x = \frac{2L}{3} \) is in Region 2 (between the charges), but it cannot be zero because the electric fields do not cancel out there.
- \( x = 2L \) is in Region 3 (to the right of both charges), where the electric fields can cancel out.

Thus, the location on the x-axis where the net electric field is zero is:
\[ \boxed{2L} \]