Concepts Of Physics MCQ Edition [Volume 2]PhysicsElectric Field and Potential
At a distance of 40 cm from a point charge, the electric field has a magnitude of 5.0 N C ⁻¹ . Determine the magnitude of the charge.
Options
- A2.2 10⁻¹⁰ C
- B8.9 10⁻⁷ C
- C8.9 10⁻⁹ C
- D8.9 10⁻¹¹ C
Correct answer
D. 8.9 10⁻¹¹ C
Step-by-step solution
The electric field due to a point charge q at a distance r is given by E = 1 4 ₀ q r^2 . Given E = 5.0 N C ⁻¹ and r = 40 cm = 0.4 m . Substituting the values: 5.0 = 9 10^9 q (0.4)^2 q = 5.0 0.16 9 10^9 q = 0.8 9 10^9 q = 8.88 10⁻¹¹ C 8.9 10⁻¹¹ C