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Step-by-Step Solution
Step 1: Label the Square and Charges
Consider a square ABCD of side length $a$. Let corners A and C each have a charge $Q$, and corners B and D each have a charge $q$. We are interested in the net electrostatic force on the charge at corner A.
Step 2: Identify Distances Between Charges
The distance between nearest corners (e.g., A and B, or A and D) is $a$.
The distance between opposite corners (A and C) along the diagonal is $ \sqrt{2}\,a $.
Step 3: Express the Forces Acting on Charge $Q$ at A
Three charges exert forces on $Q$ placed at A:
Charge $q$ at B: Force magnitude $F_{AB} = k \frac{Q\,q}{a^2}$, directed along AB.
Charge $q$ at D: Force magnitude $F_{AD} = k \frac{Q\,q}{a^2}$, directed along AD.
Charge $Q$ at C: Force magnitude $F_{AC} = k \frac{Q\,Q}{(\sqrt{2} a)^2} = k \frac{Q^2}{2 a^2}$, directed along AC.
Here, $k$ is the Coulomb constant (often written as $k = \frac{1}{4\pi\epsilon_0}$).
Step 4: Resolve Forces into Components
Because the forces from charges at B and D are each along perpendicular sides of the square, their vector sum can be broken down as follows:
The force from $q$ at B acts horizontally (to the right if we assume A is at the bottom left).
The force from $q$ at D acts vertically upwards (if D is above A).
The diagonal force due to the charge $Q$ at C will act along the diagonal from A to C. If $Q$ and $q$ have opposite signs, the direction of each individual force might be attractive or repulsive, but we focus on magnitudes and directions carefully to achieve net zero force.
Step 5: Condition for Net Force on $Q$ at A to be Zero
For the net force at A to be zero, the vector sum of the forces from B and D must cancel the force from C. In a symmetric square arrangement, this implies:
$$
\sqrt{\left(F_{AB}\right)^2 + \left(F_{AD}\right)^2} = F_{AC}
$$
Because $F_{AB} = k \frac{Q\,q}{a^2}$ and $F_{AD} = k \frac{Q\,q}{a^2}$, their resultant in magnitude (by perpendicular vector addition) is
$$
F_{\text{BD-resultant}}
= \sqrt{ \left( k \frac{Q\,q}{a^2} \right)^2 + \left( k \frac{Q\,q}{a^2} \right)^2 }
= k \frac{Q\,q}{a^2} \sqrt{2}.
$$
Thus, we set this equal to the magnitude of the force due to the charge at C:
$$
k \frac{Q\,q}{a^2} \sqrt{2} = k \frac{Q^2}{2 a^2}.
$$
Step 6: Solve for $ \frac{Q}{q} $
Canceling the common factor $k / a^2$ from both sides, we get:
$$
Q\, q \,\sqrt{2} = \frac{Q^2}{2}.
$$
Divide by $Q \neq 0$:
$$
q \,\sqrt{2} = \frac{Q}{2}.
$$
Rearrange for $ \frac{Q}{q} $:
$$
\frac{Q}{q} = 2 \sqrt{2}.
$$
However, the problem's given correct answer is $ - 2 \sqrt{2} $. This implies that the charges must be opposite in sign (e.g., $Q$ is negative while $q$ is positive, or vice versa). The magnitude of the ratio is $2 \sqrt{2}$, and the negative sign indicates the difference in charge signs.
Step 7: Conclude the Ratio
Therefore, considering the sign convention for net force to be zero (resultant directions must cancel), we have:
$$
\frac{Q}{q} = -2\sqrt{2}.
$$
The negative sign shows that $Q$ and $q$ are oppositely charged.
Final Answer
$$
\boxed{\frac{Q}{q} = - 2 \sqrt{2}}.
$$
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