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 that this body may be in equilibrium?

Options

A. stable equilibrium
B. unstable equilibrium
C. neutral equilibrium
D. cannot be determined

Explanation

To find the value of charge \( Q \) that allows a small body of mass \( m \) with a unit positive charge to be in equilibrium above the center of a square formed by four equal positive charges at its corners, we can follow these steps:

### Step 1: Understanding the Configuration
We have four charges \( Q \) placed at the corners of a square. The diagonal of the square is given as \( 2a \). The distance \( a \) can be derived from the diagonal using the relation:
\[ a = \frac{2a}{\sqrt{2}} = a\sqrt{2} \]
Thus, the side length \( s \) of the square is:
\[ s = \frac{2a}{\sqrt{2}} = a\sqrt{2} \]

### Step 2: Finding the Distance from Charges to the Body
The small body is placed at a height \( h \) above the center of the square. The distance from the center of the square to any corner (where the charge \( Q \) is located) can be calculated as:
\[ d = \sqrt{\left(\frac{s}{2}\right)^2 + h^2} = \sqrt{\left(\frac{a\sqrt{2}}{2}\right)^2 + h^2} = \sqrt{\frac{a^2}{2} + h^2} \]

### Step 3: Calculating the Electric Force on the Body
The electric force \( F \) acting on the small body due to one charge \( Q \) is given by Coulomb's law:
\[ F = k \frac{Q \cdot 1}{d^2} = k \frac{Q}{\left(\frac{a^2}{2} + h^2\right)} \]
where \( k \) is Coulomb's constant.

### Step 4: Resolving Forces
Since there are four charges, we need to consider the net force acting on the body. The forces due to the four charges will have both horizontal and vertical components. The horizontal components will cancel out due to symmetry, while the vertical components will add up.

The vertical component of the force due to one charge is:
\[ F_{vertical} = F \cdot \frac{h}{d} = k \frac{Q}{\left(\frac{a^2}{2} + h^2\right)} \cdot \frac{h}{\sqrt{\frac{a^2}{2} + h^2}} \]
Thus, for four charges, the total vertical force \( F_{net} \) is:
\[ F_{net} = 4 \cdot k \frac{Q}{\left(\frac{a^2}{2} + h^2\right)} \cdot \frac{h}{\sqrt{\frac{a^2}{2} + h^2}} \]

### Step 5: Setting Up the Equilibrium Condition
For the body to be in equilibrium, the total vertical force must balance the weight of the body:
\[ F_{net} = mg \]
Substituting the expression for \( F_{net} \):
\[ 4 \cdot k \frac{Qh}{\left(\frac{a^2}{2} + h^2\right)^{3/2}} = mg \]

### Step 6: Solving for \( Q \)
Rearranging the equation to solve for \( Q \):
\[ Q = \frac{mg \left(\frac{a^2}{2} + h^2\right)^{3/2}}{4k h} \]

### Final Answer
Thus, the value of \( Q \) required for the body to be in equilibrium is:
\[ Q = \frac{mg \left(\frac{a^2}{2} + h^2\right)^{3/2}}{4k h} \]