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’s a step-by-step breakdown of the solution:

### Step 1: Understand the Configuration
We have a square with two equal charges \( Q = +\sqrt{2} \, \mu C \) placed at two opposite corners. The other two corners have equal charges \( q \). We need to find \( q \) such that the net force on \( Q \) is zero.

### Step 2: Identify Forces Acting on Charge \( Q \)
The forces acting on charge \( Q \) are:
- The attractive forces due to the charges \( q \) at the other two corners.
- The repulsive forces due to the other charge \( Q \).

### Step 3: Calculate the Distance Between Charges
Let the side length of the square be \( a \). The distance between the charges \( Q \) and \( q \) is \( a \), and the distance between the two charges \( Q \) is \( a\sqrt{2} \) (the diagonal of the square).

### Step 4: Write the Expression for Forces
The force between charge \( Q \) and charge \( q \) is given by Coulomb's law:
\[ F_{Qq} = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q \cdot q}{a^2} \]
The force between the two charges \( Q \) is:
\[ F_{QQ} = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q^2}{(a\sqrt{2})^2} = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q^2}{2a^2} \]

### Step 5: Set Up the Force Balance
Since the forces due to the two charges \( q \) must balance the repulsive force from the other charge \( Q \), we can express this as:
\[ 2F_{Qq} = F_{QQ} \]
Substituting the expressions for the forces:
\[ 2 \left( \frac{1}{4\pi \epsilon_0} \cdot \frac{Q \cdot q}{a^2} \right) = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q^2}{2a^2} \]

### Step 6: Simplify the Equation
Cancel \( \frac{1}{4\pi \epsilon_0} \) and \( a^2 \) from both sides:
\[ 2Qq = \frac{Q^2}{2} \]

### Step 7: Solve for \( q \)
Rearranging gives:
\[ q = \frac{Q}{4} \]

### Step 8: Substitute the Value of \( Q \)
Substituting \( Q = \sqrt{2} \, \mu C \):
\[ q = \frac{\sqrt{2} \, \mu C}{4} = \frac{\sqrt{2}}{4} \, \mu C \]

### Step 9: Calculate the Final Value of \( q \)
To express \( q \) in a more standard form:
\[ q = \frac{\sqrt{2}}{4} \, \mu C = \frac{1}{2\sqrt{2}} \, \mu C \approx 0.3535 \, \mu C \]

### Final Answer
Thus, the value of \( q \) that makes the resultant force on \( Q \) zero is:
\[ q = 0.3535 \, \mu C \]