Question

Three point charges are located on a circular arc at A, B and C as shown in the figure below. The total electric field at the centre of the arc (C) is

• A. 15000 N C\(^{-1}\)
• B. 10000 N C\(^{-1}\)
• C. 20000 N C\(^{-1}\)
• D. 5000 N C\(^{-1}\)

Hint:

To solve problems involving electric fields from multiple charges, always break down the field components into radial and tangential directions, then add them vectorially to find the resultant field.

Answer 1

The Correct Option is D

The electric field due to a point charge is given by Coulomb's law: \[ E = \frac{k |q|}{r^2} \] Where: 
- \( E \) is the electric field, 
- \( k = 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2} \) is Coulomb's constant, 
- \( q \) is the charge, - \( r \) is the distance from the charge to the point where the electric field is being calculated. 
Since the charges are located on a circular arc, the electric field at the center due to each charge will have both a radial and tangential component.

The components from each charge can be added vectorially to find the resultant electric field. Given: - \( q_A = +4.0 \, \text{nC} \), - \( q_B = +4.0 \, \text{nC} \), - \( q_C = -2.0 \, \text{nC} \), - \( r = 6.0 \, \text{cm} = 0.06 \, \text{m} \), - The angle between the charges is \( 60^\circ \). Since the field vectors due to the charges \( A \) and \( B \) will cancel out in the tangential direction and add up in the radial direction, and the field due to charge \( C \) contributes directly along the axis, the net electric field at the center will be the resultant of these components. 
After vector addition, the magnitude of the net electric field is approximately \( 5000 \, \text{N C}^{-1} \).