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 find the ratio of the electric field intensities \( \frac{E_1}{E_2} \) at the corners and midpoints of the sides of a square due to a charge \( Q \) placed at the center, we can follow these steps:

### Step-by-Step Solution:

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

2. **Determine Distances**:
- The distance from the center of the square to a corner of the square is given by the formula for the diagonal of a square:
\[ \text{Diagonal} = \sqrt{2} \cdot a \]
Therefore, the distance from the center to a corner (which is half the diagonal) is:
\[ r_1 = \frac{\sqrt{2} \cdot a}{2} = \frac{a}{\sqrt{2}} \]
- The distance from the center of the square to the midpoint of a side is:
\[ r_2 = \frac{a}{2} \]

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

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

### Final Answer:
The ratio of the electric field intensities is:
\[ \frac{E_1}{E_2} = \frac{1}{2} \]