A charge Q is placed at the centre of a square If electric field intensity due to the charge at the corners of the square is `E_(1)` and the intensity at the mid point of the sidde of square is `E_(2)` then the ratio of `(E_(1))/(E_(2))` will be

Options

A. `(1)/(2sqrt(2))`
B. `sqrt(2)`
C. `(1)/(2)`
D. 2

Answer

To solve the problem, we need to find the ratio of the electric field intensities \( E_1 \) and \( E_2 \) at the corners and midpoints of the sides of a square, respectively, when a charge \( Q \) is placed at the center of the square.

### Step-by-Step Solution:

1. **Identify the Geometry**:
- Consider a square with side length \( A \).
- The charge \( Q \) is at the center of the square.

2. **Calculate the Distance to the Corners**:
- The distance from the center of the square to any corner can be calculated using the Pythagorean theorem.
- The diagonal of the square is \( \sqrt{A^2 + A^2} = A\sqrt{2} \).
- Therefore, the distance \( d_1 \) from the center to a corner is:
\[ d_1 = \frac{A\sqrt{2}}{2} = \frac{A}{\sqrt{2}} \]

3. **Calculate the Distance to the Midpoints of the Sides**:
- The distance from the center of the square to the midpoint of any side is half the side length:
- Therefore, the distance \( d_2 \) is:
\[ d_2 = \frac{A}{2} \]

4. **Calculate the Electric Field Intensities**:
- The electric field intensity \( E \) due to a point charge \( Q \) at a distance \( r \) is given by:
\[ E = k \frac{Q}{r^2} \]
- For the corners (distance \( d_1 \)):
\[ E_1 = k \frac{Q}{d_1^2} = k \frac{Q}{\left(\frac{A}{\sqrt{2}}\right)^2} = k \frac{Q}{\frac{A^2}{2}} = \frac{2kQ}{A^2} \]
- For the midpoints (distance \( d_2 \)):
\[ E_2 = k \frac{Q}{d_2^2} = k \frac{Q}{\left(\frac{A}{2}\right)^2} = k \frac{Q}{\frac{A^2}{4}} = \frac{4kQ}{A^2} \]

5. **Calculate the Ratio \( \frac{E_1}{E_2} \)**:
- Now, we can find the ratio:
\[ \frac{E_1}{E_2} = \frac{\frac{2kQ}{A^2}}{\frac{4kQ}{A^2}} = \frac{2}{4} = \frac{1}{2} \]

### Final Answer:
The ratio \( \frac{E_1}{E_2} \) is \( \frac{1}{2} \).