Question

Four point charges of 1 \(\mu C\), -2 \(\mu C\), 1 \(\mu C\), and -2 \(\mu C\) are placed at the corners A, B, C, and D respectively, of a square of side 30 cm. Find the net force acting on a charge of 4 \(\mu C\) placed at the center of the square.

Hint:

Remember to apply Coulomb’s law for each charge, and use vector addition to determine the resultant force. The forces should be broken down into their components for easier addition.

Answer 1

Given:

Charges at the corners:\(q_A = 1 \, \mu\text{C}\) 
 \(q_B = -2 \, \mu\text{C}\) 
 \(q_C = 1 \, \mu\text{C}\) 
 \(q_D = -2 \, \mu\text{C}\) 
 Side of the square, \(a = 30 \, \text{cm} = 0.3 \, \text{m}\)
Charge at the center, \(q_0 = 4 \, \mu\text{C}\) 

Step 1: Calculate the distance from the center to a corner. The distance (\(r\)) from the center of the square to any corner is half the length of the diagonal of the square. The diagonal (\(d\)) of the square is: \[ d = a\sqrt{2} = 0.3 \times \sqrt{2} \, \text{m} \] Thus, the distance from the center to a corner is:
\[ r = \frac{d}{2} = \frac{0.3 \times \sqrt{2}}{2} = 0.15 \times \sqrt{2} \, \text{m} \] 

Step 2: Calculate the force due to each charge.
The force (\(F\)) between two charges is given by Coulomb's law:
\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \] where \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\). 

Forces due to charges at corners A, B, C, and D:
1. **Force due to \(q_A\)**:
\[ F_A = \frac{9 \times 10^9 \cdot |1 \times 10^{-6} \cdot 4 \times 10^{-6}|}{(0.15 \times \sqrt{2})^2} \] \[ F_A = \frac{36 \times 10^{-3}}{0.045} = 0.8 \, \text{N} \] The direction of \(F_A\) is along the line from A to the center. 2. **Force due to \(q_B\)**:
\[ F_B = \frac{9 \times 10^9 \cdot |{-2} \times 10^{-6} \cdot 4 \times 10^{-6}|}{(0.15 \times \sqrt{2})^2} \] \[ F_B = \frac{72 \times 10^{-3}}{0.045} = 1.6 \, \text{N} \] The direction of \(F_B\) is along the line from B to the center. 3. **Force due to \(q_C\)**:
\[ F_C = \frac{9 \times 10^9 \cdot |1 \times 10^{-6} \cdot 4 \times 10^{-6}|}{(0.15 \times \sqrt{2})^2} \] \[ F_C = \frac{36 \times 10^{-3}}{0.045} = 0.8 \, \text{N} \] The direction of \(F_C\) is along the line from C to the center. 4. **Force due to \(q_D\)**:
\[ F_D = \frac{9 \times 10^9 \cdot |{-2} \times 10^{-6} \cdot 4 \times 10^{-6}|}{(0.15 \times \sqrt{2})^2} \] \[ F_D = \frac{72 \times 10^{-3}}{0.045} = 1.6 \, \text{N} \] The direction of \(F_D\) is along the line from D to the center. Step 3: Resolve forces into components.
The forces \(F_A\) and \(F_C\) are along the diagonals of the square, and \(F_B\) and \(F_D\) are along the other diagonals. Due to symmetry, the horizontal and vertical components of the forces cancel out. Step 4: Calculate the net force.
Since the forces are symmetrically distributed and their components cancel out, the **net force on the charge at the center is zero**. Final Answer: \[ \boxed{\text{The net force acting on the charge at the center is } 0 \, \text{N.}} \]

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