Question

Two charges 14 µC and −4 µC are placed at (−12 cm, 0, 0) and (12 cm, 0, 0) in an external electric field `E = (B/r^2)`, where B = 1.2 × 10⁶ N/(cm²) and r is in metres. The electrostatic potential energy of the configuration is ______.

पर्याय

• 97.9 J

• 102.1 J

• 2.1 J

• −97.9 J

Answer 1

Two charges 14 µC and −4 µC are placed at (−12 cm, 0, 0) and (12 cm, 0, 0) in an external electric field `E = (B/r^2)`, where B = 1.2 × 10⁶ N/(cm²) and r is in metres. The electrostatic potential energy of the configuration is 97.9 J.

Explanation:

Energy = q₁V + q₂V + kq₁q₂/r

E = `B/r^2`

∴ `V = intB/r^2dr`

= `B/r`

= 1.2 × 10⁶/r

Now, energy = 14 × 10⁻⁶ × 1.2 × 10⁶/(12 × 10⁻²) −4 × 10⁻⁶ × 1.2 × 10⁶/(1.2 × 10⁻²) −9 × 10⁹ × 14 ×10⁻⁶ × 4 × 10⁻⁶/(24 × 10⁻²)

= (14 − 4) × 1.2/(12 × 10⁻²) − 9 × 56/240

= 100 − 2.1

= 97.9 J