PhysicsCBSE Class 12 · Board exam 2026, set 55/1/1 · 3 marks · Short answer

CBSE Class 12 Physics 2026: question 24

Set in the CBSE Class 12 board exam of 2026, paper 55/1/1, section C, question 24, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q24 · 3 marks · Short answerElectrostatic Potential and Capacitance

Figure shows a narrow beam of electrons entering with a velocity of $3 \times 1 0^{7}$ m/s, symmetrically through the space between two parallel horizontal plates $P_{1} P_{1}^{'}$ and $P_{2} P_{2}^{'}$ kept 2 cm apart.
If each plate is 3 cm long, calculate the potential difference V applied between the plates so that the beam just strikes the end $P_{2}^{'}$ .

Figure in the paper: Two parallel horizontal plates P1P1' (top, with potential difference V applied) and P2P2' (bottom). An electron (charge -e, mass m) enters midway between the plates moving horizontally to the right; its path curves downward and just reaches the end P2' of the lower plate. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/1/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

The motion is like a projectile: uniform horizontal velocity and uniform vertical acceleration.

Hint 2

The beam starts midway, so its vertical deflection is half the plate separation. Relate the acceleration to $e V / (m d)$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

The electron enters midway, so to just strike $P_{2}^{'}$ it must be deflected by $y = \frac{d}{2} = 1$ cm $= 0.01$ m while travelling $x = 3$ cm $= 0.03$ m horizontally.
½ mark
Time between the plates: $t = \frac{x}{v} = \frac{0.03}{3 \times 1 0 ^{7}} = 1 \times 1 0^{- 9}$ s
½ mark
From $y = \frac{1}{2} a t^{2}$ : $a = \frac{2 y}{t ^{2}} = \frac{2 \times 0.01}{( 1 0 ^{- 9} ) ^{2}} = 2 \times 1 0^{16}$ m/s²
½ mark
The acceleration comes from the field $E = \frac{V}{d}$ : $a = \frac{e E}{m} = \frac{e V}{m d}$
½ mark
$V = \frac{am d}{e} = \frac{2 \times 1 0 ^{16} \times 9.1 \times 1 0 ^{- 31} \times 0.02}{1.6 \times 1 0 ^{- 19}} \approx 2.3 \times 1 0^{3}$ V
1 mark

Answer: V ≈ 2.3 × 10³ V (about 2275 V)

Where the marks are

Takes the vertical deflection as 1 cm over a length of 3 cm (½ mark)
Time of flight 10⁻⁹ s (½ mark)
Acceleration 2 × 10¹⁶ m/s² (½ mark)
Uses a = eV/md (½ mark)
V ≈ 2.3 kV (1 mark)

Revise the chapter

Chapter: Electrostatic Potential and Capacitance (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Electrostatic Potential and Capacitance, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 55/1/1: every question, by section.

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CBSE Class 12 Physics 2026: question 24

Hints

Step-by-step solution

Revise the chapter

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