There are four charges, each with a magnitude of `2.0 mu C`. Two are positive and two are negative. The charges are fixed to the corners of a 0.30 m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.

Options

A. 0.20 N
B. 0.57 N
C. 0.37 N
D. 0.16 N

Answer

To find the magnitude of the net electrostatic force experienced by any charge in the given configuration, we can follow these steps:

### Step 1: Understanding the Configuration
We have four charges arranged at the corners of a square with a side length of 0.30 m. The charges are:
- Two positive charges (+2 µC each)
- Two negative charges (-2 µC each)

The configuration is such that the net force on any charge is directed toward the center of the square.

### Step 2: Identify the Forces Acting on One Charge
Let's consider one of the positive charges located at one corner of the square. The forces acting on this charge are due to the two negative charges located at the opposite corners.

### Step 3: Calculate the Distance Between Charges
The distance between the positive charge and each negative charge is the diagonal of the square. The diagonal \(d\) of a square with side length \(a\) is given by:
\[ d = a\sqrt{2} = 0.30\sqrt{2} \text{ m} \]

### Step 4: Calculate the Electrostatic Force Between Charges
Using Coulomb's Law, the force \(F\) between two point charges is given by:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where:
- \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\) (Coulomb's constant)
- \(q_1 = q_2 = 2 \times 10^{-6} \, \text{C}\)
- \(r = 0.30 \, \text{m}\) (for the charges on the same side)
- \(r = 0.30\sqrt{2} \, \text{m}\) (for the diagonal)

#### Force between a positive charge and a negative charge on the same side:
\[ F_1 = k \frac{(2 \times 10^{-6})^2}{(0.30)^2} = 9 \times 10^9 \frac{4 \times 10^{-12}}{0.09} = 0.4 \, \text{N} \]

#### Force between a positive charge and a negative charge on the diagonal:
\[ F_2 = k \frac{(2 \times 10^{-6})^2}{(0.30\sqrt{2})^2} = 9 \times 10^9 \frac{4 \times 10^{-12}}{0.18} = 0.2 \, \text{N} \]

### Step 5: Calculate the Resultant Force
The net force on the positive charge is the vector sum of the forces \(F_1\) and \(F_2\). Since \(F_1\) is directed towards one negative charge and \(F_2\) towards the other, we can use the Pythagorean theorem to find the resultant force \(F_R\):
\[ F_R = \sqrt{F_1^2 + F_2^2} = \sqrt{(0.4)^2 + (0.2)^2} = \sqrt{0.16 + 0.04} = \sqrt{0.2} \approx 0.447 \, \text{N} \]

### Step 6: Conclusion
The magnitude of the net electrostatic force experienced by any charge is approximately \(0.447 \, \text{N}\).

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