Physics · Electrostatics · chapter 1 of 14
Class 12 Physics Chapter 1: Electric Charges and Fields case study question with answers
One case-based question of the kind the board asks: a passage and a figure, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
Case study · 4 marks
In a laboratory, two horizontal metal plates are connected to a supply so that a uniform electric field of 4.0 × 10⁴ N C⁻¹ acts in the space between them (see figure). A fine spray puts tiny oil drops into this space, and some drops pick up electrons from the air. A student watches one drop of mass 3.2 × 10⁻¹⁵ kg through a microscope and adjusts the field until the drop stays still, neither rising nor falling. (Take g = 10 m s⁻² and e = 1.6 × 10⁻¹⁹ C.)
Part 1 · 1 mark
The drop carries extra electrons. For it to stay still, the electric field between the plates must point
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Answer: (a) vertically downward
The electric force must balance mg, so it must act upward. On a negative charge F = qE points opposite to E, so E must point downward.
Part 2 · 1 mark
The magnitude of the charge on the drop is
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Answer: (a) 8.0 × 10⁻¹⁹ C
qE = mg, so q = mg/E = 3.2 × 10⁻¹⁵ × 10/(4.0 × 10⁴) = 8.0 × 10⁻¹⁹ C.
Part 3 · 2 marks
How many excess electrons does the drop carry? Explain why the answer must be a whole number.
Answer and marking scheme
n = q/e = 8.0 × 10⁻¹⁹/1.6 × 10⁻¹⁹ = 5. Charge is quantised: every free charge is an integral multiple of e (q = ne), so a drop can carry only a whole number of electrons.
- n = q/e = 5 (1)
- charge is quantised, q = ne with n an integer (1)
Or
With the drop at rest, the field is suddenly doubled in magnitude, keeping its direction. Find the acceleration of the drop, with its direction.
Answer and marking scheme
Electric force becomes 2qE = 2mg, upward; weight mg downward. Net force = mg upward, so a = g = 10 m s⁻², vertically upward.
- electric force doubles to 2mg, net force mg (1)
- a = 10 m s⁻², upward (1)