Two equal point charges `Q=+sqrt(2) mu C` are placed at each of the two opposite corners of a square and equal point charges q at each of the other two corners. What must be the value of q so that the resultant force on Q is zero?

Explanation

To solve the problem, we need to determine the value of charge \( q \) such that the resultant force on charge \( Q \) is zero. Here are the steps to find the solution:

### Step 1: Understand the Configuration
We have two equal charges \( Q = +\sqrt{2} \, \mu C \) placed at two opposite corners of a square. Let's denote these charges as \( Q_1 \) and \( Q_2 \). The other two corners of the square have equal charges \( q \).

### Step 2: Identify Forces Acting on Charge \( Q \)
The forces acting on charge \( Q \) due to the other charges are:
- The repulsive force due to the other charge \( Q \) (let's denote it as \( F_{QQ} \)).
- The attractive or repulsive force due to the charges \( q \) (let's denote these forces as \( F_{Qq1} \) and \( F_{Qq2} \)).

### Step 3: Calculate the Force Between Charges \( Q \)
The distance between the two charges \( Q \) is equal to the diagonal of the square. If the side of the square is \( a \), then the distance \( d \) between the two charges \( Q \) is given by:
\[ d = \sqrt{2} a \]
Using Coulomb's law, the force \( F_{QQ} \) between the charges \( Q_1 \) and \( Q_2 \) is:
\[ F_{QQ} = k \frac{Q^2}{d^2} = k \frac{Q^2}{2a^2} \]

### Step 4: Calculate the Force Between Charge \( Q \) and Charge \( q \)
The distance between charge \( Q \) and charge \( q \) is \( a \). The force \( F_{Qq} \) between charge \( Q \) and charge \( q \) is:
\[ F_{Qq} = k \frac{Q |q|}{a^2} \]

### Step 5: Set Up the Force Balance
For the net force on charge \( Q \) to be zero, the forces acting on it must balance. Since \( F_{QQ} \) is repulsive and \( F_{Qq} \) is attractive (if \( q \) is negative), we can write:
\[ F_{QQ} = 2 F_{Qq} \]
This is because there are two charges \( q \) acting on \( Q \).

### Step 6: Substitute the Forces
Substituting the expressions for the forces, we have:
\[ k \frac{Q^2}{2a^2} = 2 \left( k \frac{Q |q|}{a^2} \right) \]
Cancelling \( k \) and \( a^2 \) from both sides gives:
\[ \frac{Q^2}{2} = 2 |q| \implies |q| = \frac{Q^2}{4} \]

### Step 7: Substitute the Value of \( Q \)
Substituting \( Q = \sqrt{2} \, \mu C \):
\[ |q| = \frac{(\sqrt{2} \, \mu C)^2}{4} = \frac{2 \, \mu C^2}{4} = \frac{1}{2} \, \mu C^2 \]

### Step 8: Determine the Sign of \( q \)
Since \( q \) must be negative to attract \( Q \) and balance the repulsive force from the other \( Q \), we conclude:
\[ q = -\frac{1}{2} \, \mu C \]

### Final Answer
Thus, the value of charge \( q \) must be:
\[ \boxed{-\frac{1}{2} \, \mu C} \]

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