Four points charges -q,+q,+q and -q are placed on y-axis at y=-2d, y=-d,y=+d and y=+2d, respectively. The magnitude of the electric field E at a point on the x-axis at x=D, with `Dgtgtd`, will behave as :

Options

A. `Eprop1/(D^(3))`
B. `E prop 1/D`
C. `E prop1/(D^(4))`
D. `E prop1/(D^(2))`

Answer

To solve the problem, we need to calculate the electric field \( E \) at a point on the x-axis at \( x = D \) due to the four point charges located on the y-axis. The charges are positioned at \( y = -2d \) (charge \( -q \)), \( y = -d \) (charge \( +q \)), \( y = +d \) (charge \( +q \)), and \( y = +2d \) (charge \( -q \)).

### Step 1: Identify the distances from the point on the x-axis to each charge

The point on the x-axis is at \( (D, 0) \). The distances from this point to each charge can be calculated using the Pythagorean theorem:

1. For the charge at \( y = -2d \):
\[ r_1 = \sqrt{D^2 + (2d)^2} = \sqrt{D^2 + 4d^2} \]

2. For the charge at \( y = -d \):
\[ r_2 = \sqrt{D^2 + d^2} \]

3. For the charge at \( y = +d \):
\[ r_3 = \sqrt{D^2 + d^2} \]

4. For the charge at \( y = +2d \):
\[ r_4 = \sqrt{D^2 + 4d^2} \]

### Step 2: Calculate the electric field due to each charge

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

1. Electric field due to the charge at \( y = -2d \) (charge \( -q \)):
\[ E_1 = \frac{kq}{r_1^2} = \frac{kq}{D^2 + 4d^2} \]

2. Electric field due to the charge at \( y = -d \) (charge \( +q \)):
\[ E_2 = \frac{kq}{r_2^2} = \frac{kq}{D^2 + d^2} \]

3. Electric field due to the charge at \( y = +d \) (charge \( +q \)):
\[ E_3 = \frac{kq}{r_3^2} = \frac{kq}{D^2 + d^2} \]

4. Electric field due to the charge at \( y = +2d \) (charge \( -q \)):
\[ E_4 = \frac{kq}{r_4^2} = \frac{kq}{D^2 + 4d^2} \]

### Step 3: Determine the direction of the electric fields

- The electric field due to the negative charges (\( -q \)) will point towards the charge.
- The electric field due to the positive charges (\( +q \)) will point away from the charge.

### Step 4: Calculate the net electric field at point \( (D, 0) \)

The net electric field \( E \) at point \( (D, 0) \) is the vector sum of the individual electric fields. Since the electric fields due to \( +q \) charges at \( y = -d \) and \( y = +d \) are in the same direction (away from the charges), and the electric fields due to \( -q \) charges at \( y = -2d \) and \( y = +2d \) are in the opposite direction, we can express the net electric field as:

\[ E_{\text{net}} = E_2 + E_3 - E_1 - E_4 \]

Substituting the expressions for \( E_1, E_2, E_3, \) and \( E_4 \):

\[ E_{\text{net}} = \frac{kq}{D^2 + d^2} + \frac{kq}{D^2 + d^2} - \frac{kq}{D^2 + 4d^2} - \frac{kq}{D^2 + 4d^2} \]

This simplifies to:

\[ E_{\text{net}} = 2\frac{kq}{D^2 + d^2} - 2\frac{kq}{D^2 + 4d^2} \]

### Step 5: Factor out common terms

Factoring out \( 2kq \):

\[ E_{\text{net}} = 2kq \left( \frac{1}{D^2 + d^2} - \frac{1}{D^2 + 4d^2} \right) \]

### Step 6: Simplify the expression

To combine the fractions:

\[ E_{\text{net}} = 2kq \left( \frac{(D^2 + 4d^2) - (D^2 + d^2)}{(D^2 + d^2)(D^2 + 4d^2)} \right) \]

This simplifies to:

\[ E_{\text{net}} = 2kq \left( \frac{3d^2}{(D^2 + d^2)(D^2 + 4d^2)} \right) \]

### Step 7: Analyze the behavior as \( D \gg d \)

As \( D \) becomes much larger than \( d \), we can approximate:

\[ E_{\text{net}} \approx \frac{6kqd^2}{D^4} \]

Thus, the electric field \( E \) behaves as:

\[ E \propto \frac{1}{D^4} \]

### Conclusion

The magnitude of the electric field \( E \) at the point on the x-axis at \( x = D \) behaves as inversely proportional to \( D^4 \).