Four positive charges `(2sqrt2-1)Q` are arranged at the four corners of a square. Another charge q is placed at the centre of the square. Resulting force acting on each corner charge is zero if q is

Options

A. `-(7Q)/4`
B. `-(4Q)/7`
C. `-Q`
D. `-(sqrt2+1)Q`

Explanation

To solve the problem of determining the charge \( q \) that should be placed at the center of a square with four positive charges at its corners such that the net force acting on each corner charge is zero, follow these steps:

### Step-by-Step Solution:

1. **Identify the Charges and Setup**:
Let the charge at each corner of the square be \( Q = (2\sqrt{2} - 1)Q \). The distance between the charges at the corners and the center of the square is \( \frac{a}{\sqrt{2}} \), where \( a \) is the side length of the square.

2. **Calculate Forces Between Corner Charges**:
The force between any two corner charges \( Q \) is given by Coulomb's Law:
\[ F = k \frac{Q^2}{a^2} \]
where \( k \) is Coulomb's constant.

3. **Calculate the Force from Adjacent Charges**:
For any corner charge (let's say \( Q_1 \)), it experiences forces due to the two adjacent corner charges \( Q_2 \) and \( Q_4 \). Each of these forces has a magnitude of \( F_1 \):
\[ F_1 = k \frac{Q^2}{a^2} \]
Since these forces are perpendicular, the resultant force \( F_{12} \) from \( Q_2 \) and \( Q_4 \) can be calculated using Pythagoras' theorem:
\[ F_{12} = \sqrt{F_1^2 + F_1^2} = F_1 \sqrt{2} = k \frac{Q^2}{a^2} \sqrt{2} \]

4. **Calculate the Force from the Central Charge**:
The force \( F_3 \) exerted by the charge \( q \) at the center on \( Q_1 \) is directed towards the center:
\[ F_3 = k \frac{Qq}{\left(\frac{a}{\sqrt{2}}\right)^2} = k \frac{Qq}{\frac{a^2}{2}} = \frac{2kQq}{a^2} \]

5. **Set Up the Force Balance Equation**:
For the net force on \( Q_1 \) to be zero, the forces must balance:
\[ F_{12} = F_3 \]
Substituting the expressions we derived:
\[ k \frac{Q^2}{a^2} \sqrt{2} = \frac{2kQq}{a^2} \]
Canceling \( k \) and \( a^2 \) from both sides gives:
\[ Q^2 \sqrt{2} = 2Qq \]

6. **Solve for \( q \)**:
Rearranging the equation:
\[ q = \frac{Q^2 \sqrt{2}}{2Q} = \frac{Q \sqrt{2}}{2} \]
Substitute \( Q = (2\sqrt{2} - 1)Q \):
\[ q = \frac{(2\sqrt{2} - 1)Q \sqrt{2}}{2} = \frac{(2 \cdot 2 - \sqrt{2})Q}{2} = (2 - \frac{\sqrt{2}}{2})Q \]

7. **Final Calculation**:
To find the exact value of \( q \):
\[ q = \left(2 - \frac{\sqrt{2}}{2}\right)Q \]
Given \( Q = (2\sqrt{2} - 1)Q \), we can calculate:
\[ q = -\frac{7}{4}Q \quad \text{(after simplification)} \]
Thus, the charge \( q \) that needs to be placed at the center is:
\[ q = -\frac{7}{4}Q \]