Concepts Of Physics MCQ Edition [Volume 2]PhysicsElectric Field and Potential
As shown in the figure, an electric field of magnitude 1000 ext N C ⁻¹ is established between two parallel plates separated by a distance of 2.0 ext cm . Evaluate the following: (i) The potential difference between the two plates. (ii) The minimum speed at which an electron must be projected from the lower plate in the direction of the field in order to reach the upper plate. (iii) The maximum height attained by the
Options
- A(i) 20 V , (ii) 1.87 10^6 m s ⁻¹ , (iii) 0.25 cm
- B(i) 40 V , (ii) 3.75 10^6 m s ⁻¹ , (iii) 0.50 cm
- C(i) 20 V , (ii) 2.65 10^6 m s ⁻¹ , (iii) 1.50 cm
- D(i) 20 V , (ii) 2.65 10^6 m s ⁻¹ , (iii) 0.50 cm
Correct answer
D. (i) 20 V , (ii) 2.65 10^6 m s ⁻¹ , (iii) 0.50 cm
Step-by-step solution
Given: Electric field, E = 1000 N C ⁻¹ Distance between plates, d = 2.0 cm = 0.02 m Charge of an electron, e = 1.6 10⁻¹⁹ C Mass of an electron, m = 9.1 10⁻³¹ kg (i) The potential difference V between the two plates is given by: V = E d V = 1000 0.02 = 20 V (ii) The electron is projected from the lower plate in the direction of the electric field (upwards). The electric force on the electron is downwards, so it experiences a retardation a = eE m . For the electron to just reach the upper plate, its final velocity at