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, we need to find the electric field at point P due to four identical charges Q located at the corners of a square with side length a. The point P is located symmetrically at a distance \( \frac{a}{\sqrt{2}} \) from the center of the square.

### Step 1: Understanding the Configuration
We have four identical charges, \( Q \), fixed at the corners of a square. Let's label the corners of the square as A, B, C, and D. The center of the square is O, and point P is located at a distance \( \frac{a}{\sqrt{2}} \) from O.

### Step 2: Calculate the Distance from Each Charge to Point P
The distance from the center O to each corner (A, B, C, D) of the square is given by:
\[ d = \frac{a}{\sqrt{2}} \]
The distance from each charge to point P can be calculated using the Pythagorean theorem. The distance from each charge to point P is:
\[ PO = \frac{a}{\sqrt{2}} \quad \text{(distance from center to P)} \]
Using the right triangle formed by the center O and the corner A, we find:
\[ PA = \sqrt{PO^2 + OA^2} = \sqrt{\left(\frac{a}{\sqrt{2}}\right)^2 + \left(\frac{a}{2}\right)^2} = \sqrt{\frac{a^2}{2} + \frac{a^2}{4}} = \sqrt{\frac{3a^2}{4}} = \frac{a\sqrt{3}}{2} \]

### Step 3: Calculate the Electric Field due to One Charge
The electric field \( E \) due to one charge \( Q \) at point P is given by Coulomb's law:
\[ E = \frac{kQ}{r^2} \]
where \( r \) is the distance from the charge to point P. Therefore:
\[ E = \frac{kQ}{\left(\frac{a\sqrt{3}}{2}\right)^2} = \frac{kQ}{\frac{3a^2}{4}} = \frac{4kQ}{3a^2} \]

### Step 4: Resolve the Electric Field into Components
Since the charges are symmetrically placed, we can resolve the electric field \( E \) into horizontal and vertical components. The angle \( \theta \) between the line connecting the charge to point P and the horizontal axis can be found using the geometry of the square:
\[ \theta = 45^\circ \]
Thus, the horizontal and vertical components of the electric field due to one charge are:
\[ E_x = E \cos(45^\circ) = \frac{4kQ}{3a^2} \cdot \frac{1}{\sqrt{2}} = \frac{4kQ}{3\sqrt{2} a^2} \]
\[ E_y = E \sin(45^\circ) = \frac{4kQ}{3a^2} \cdot \frac{1}{\sqrt{2}} = \frac{4kQ}{3\sqrt{2} a^2} \]

### Step 5: Calculate the Net Electric Field
Since there are two charges contributing to the horizontal component (A and C) and two charges contributing to the vertical component (B and D), the net electric field components will be:
\[ E_{net,x} = E_x + (-E_x) = 0 \quad \text{(horizontal components cancel)} \]
\[ E_{net,y} = E_y + E_y = 2E_y = 2 \cdot \frac{4kQ}{3\sqrt{2} a^2} = \frac{8kQ}{3\sqrt{2} a^2} \]

### Final Result
The net electric field at point P is directed vertically upwards and has a magnitude of:
\[ E_{net} = \frac{8kQ}{3\sqrt{2} a^2} \]