Four identical charges Q are fixed at the four corners of a square of side a. The electric field at a point P located symmetrically at a distance `a//sqrt(2)` from the center of the square is

Options

A. `(Q)/(2sqrt(2)pi epsilon_(0)a^(2))`
B. `(Q)/(sqrt(2)pi epsilon_(0)a^(2))`
C. `(2sqrt(2)Q)/(pi epsilon_(0)a^(2))`
D. `(sqrt(2)Q)/(pi epsilon_(0)a^(2))`

Explanation

To solve the problem of finding the electric field at point P due to four identical charges Q fixed at the corners of a square of side a, we can follow these steps:

### Step 1: Understand the Geometry
We have four identical charges Q located at the corners of a square. The point P is located symmetrically at a distance of \( \frac{a}{\sqrt{2}} \) from the center of the square.

### Step 2: Determine the Distance from Each Charge to Point P
The distance from the center of the square to any corner is given by \( \frac{a}{\sqrt{2}} \). Since point P is also at a distance of \( \frac{a}{\sqrt{2}} \) from the center, we can use the Pythagorean theorem to find the distance from each charge to point P.

The distance from each charge to point P can be calculated as follows:
- The distance from the center of the square to a corner is \( \frac{a}{\sqrt{2}} \).
- The distance from the center to point P is also \( \frac{a}{\sqrt{2}} \).
- Therefore, the distance from each charge to point P is \( d = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2} = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}} \).

### Step 3: Calculate the Electric Field Due to One Charge
The electric field \( E \) due to a single charge Q at a distance \( d \) is given by:
\[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{d^2} \]
Substituting \( d = \frac{a}{\sqrt{2}} \):
\[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{\left(\frac{a}{\sqrt{2}}\right)^2} = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{\frac{a^2}{2}} = \frac{2Q}{4\pi\epsilon_0 a^2} \]

### Step 4: Determine the Direction of the Electric Fields
The electric field vectors due to each charge will point away from the charges (since they are positive). The symmetry of the square means that the horizontal components of the electric fields from opposite charges will cancel out, while the vertical components will add up.

### Step 5: Calculate the Net Electric Field
Since there are four charges, and the vertical components of the electric fields will add up, we can express the net electric field \( E_{net} \) as:
\[ E_{net} = 4 \cdot E \cdot \cos(\theta) \]
Where \( \theta \) is the angle between the line connecting the charge and point P, and the vertical line. Given that \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ E_{net} = 4 \cdot \left(\frac{2Q}{4\pi\epsilon_0 a^2}\right) \cdot \frac{1}{\sqrt{2}} = \frac{8Q}{4\pi\epsilon_0 a^2 \sqrt{2}} \]

### Step 6: Final Expression
Thus, the final expression for the electric field at point P is:
\[ E_{net} = \frac{2\sqrt{2}Q}{4\pi\epsilon_0 a^2} \]