Four equal positive charges, each of value Q, are arranged at the four corners of a square of diagonal 2a. A small body of mass m carrying a unit positive charge is placed at height h above the center of the square.
What should be the value of Q in order taht this vody may be in equilibrium?

Options

A. `pi epsilon_(0)(mg)/(2h)(h^(2)+2a^(2))^(3//2)`
B. `pi epsilon_(0)(mg)/(2h)(h^(2)+a^(2))^(3//2)`
C. `pi epsilon_(0)(2mg)/(2h)(h^(2)+2a^(2))^(3//2)`
D. `pi epsilon_(0)(mg)/(2h)(h^(2)-2a^(2))^(3//2)`

Explanation

To solve the problem step by step, we need to analyze the forces acting on the small body with a unit positive charge placed at a height \( h \) above the center of the square formed by the four equal positive charges \( Q \).

### Step 1: Understanding the Geometry
- The square has a diagonal of length \( 2a \).
- The side length \( s \) of the square can be found using the relationship between the diagonal and the side of the square:
\[ s = \frac{2a}{\sqrt{2}} = a\sqrt{2} \]
- The distance from the center of the square to any corner (where the charges are located) is:
\[ r = \frac{s}{2} \sqrt{2} = a \]

### Step 2: Calculate the Distance from the Charge to the Unit Charge
- The distance \( d \) from each charge \( Q \) at the corners to the unit charge at height \( h \) is given by:
\[ d = \sqrt{a^2 + h^2} \]

### Step 3: Calculate the Electric Force from One Charge
- The electric force \( F \) exerted by one charge \( Q \) on the unit charge is given by Coulomb's law:
\[ F = \frac{1}{4\pi \epsilon_0} \frac{Q}{d^2} = \frac{1}{4\pi \epsilon_0} \frac{Q}{a^2 + h^2} \]

### Step 4: Resolve the Force into Components
- The force has both vertical and horizontal components. The vertical component \( F_v \) (which contributes to balancing the weight of the unit charge) can be found using trigonometry:
- The angle \( \theta \) between the line connecting the charge and the unit charge and the vertical line is given by:
\[ \cos \theta = \frac{h}{\sqrt{a^2 + h^2}} \]
- Therefore, the vertical component of the force from one charge is:
\[ F_v = F \cos \theta = \frac{1}{4\pi \epsilon_0} \frac{Q}{a^2 + h^2} \cdot \frac{h}{\sqrt{a^2 + h^2}} = \frac{Qh}{4\pi \epsilon_0 (a^2 + h^2)^{3/2}} \]

### Step 5: Total Vertical Force from All Four Charges
- Since there are four charges, the total vertical force \( F_{total} \) acting on the unit charge is:
\[ F_{total} = 4 \cdot F_v = 4 \cdot \frac{Qh}{4\pi \epsilon_0 (a^2 + h^2)^{3/2}} = \frac{Qh}{\pi \epsilon_0 (a^2 + h^2)^{3/2}} \]

### Step 6: Set Up the Equilibrium Condition
- For the unit charge to be in equilibrium, the total upward force must balance the weight of the unit charge \( mg \):
\[ F_{total} = mg \]
- Therefore, we have:
\[ \frac{Qh}{\pi \epsilon_0 (a^2 + h^2)^{3/2}} = mg \]

### Step 7: Solve for \( Q \)
- Rearranging the equation gives:
\[ Q = \frac{mg \pi \epsilon_0 (a^2 + h^2)^{3/2}}{h} \]

### Final Answer
Thus, the value of \( Q \) required for the small body to be in equilibrium is:
\[ Q = \frac{mg \pi \epsilon_0 (a^2 + h^2)^{3/2}}{h} \]