PhysicsCBSE Class 12 · Board exam 2026, set 55/5/1 · 4 marks · Case-based

CBSE Class 12 Physics 2026: question 29

Set in the CBSE Class 12 board exam of 2026, paper 55/5/1, section D, question 29, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q29 · 4 marks · Case-basedCompetency-basedElectrostatic Potential and Capacitance

The electric potential (V) and electric field (E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the force per unit charge at a given point in space, whereas electric potential is a scalar quantity that represents the potential energy per unit charge at a given point in space. Electric field and electric potential are related by the equations $E_{r} = \frac{- d V}{d r}$ and $E = E_{r} \overset{r}{^}$ , i.e., electric field is the negative gradient of the electric potential. This means that electric field points in the direction of decreasing potential and its magnitude is the rate of change of potential with distance. The electric field is the force that drives a unit charge to move from higher potential region to lower potential region and electric potential difference between the two points determines the work done in moving a unit charge from one point to the other point.

A pair of square conducting plates having sides of length 0·05 m are arranged parallel to each other in x-y plane. They are 0·01 m apart along z-axis and are connected to a 200 V power supply as shown in the figure. An electron enters with a speed of $3 \times 1 0^{7} ms^{- 1}$ horizontally and symmetrically in the space between the two plates. Neglect the effect of gravity on the electron.

(i)
The electric field $E$ in the region between the plates is :
1. (A) $(2 \times 1 0^{2} \frac{V}{m}) \hat{k}$
2. (B) $- (2 \times 1 0^{2} \frac{V}{m}) \hat{k}$
3. (C) $(2 \times 1 0^{4} \frac{V}{m}) \hat{k}$
4. (D) $- (2 \times 1 0^{4} \frac{V}{m}) \hat{k}$
1 mark
(ii)
In the region between the plates, the electron moves with an acceleration $a$ given by :
1. (A) $- (3 \cdot 5 \times 1 0^{15} ms^{- 2}) \hat{k}$
2. (B) $(3 \cdot 5 \times 1 0^{15} ms^{- 2}) \hat{k}$
3. (C) $(3 \cdot 5 \times 1 0^{13} ms^{- 2}) \hat{i}$
4. (D) $- (3 \cdot 5 \times 1 0^{13} ms^{- 2}) \hat{i}$
1 mark
(iii)(a)
Time interval during which an electron moves through the region between the plates is :
1. (A) $9 \cdot 0 \times 1 0^{- 9}$ s
2. (B) $1 \cdot 67 \times 1 0^{- 8}$ s
3. (C) $1 \cdot 67 \times 1 0^{- 9}$ s
4. (D) $2 \cdot 17 \times 1 0^{- 9}$ s
1 mark
OR
(iii)(b)
The vertical displacement of the electron which travels through the region between the plates is :
1. (A)10 mm
2. (B)4·9 mm
3. (C)5·9 mm
4. (D)3·0 mm
1 mark
(iv)
Which one of the following is the path traced by the electron in between the two plates ?
1. (A)a
2. (B)b
3. (C)c
4. (D)d
1 mark

Figure in the paper: Two parallel horizontal plates of length 0·05 m, separated by 0·01 m, connected to a 200 V battery (+ terminal to the upper plate, − to the lower). An electron enters midway between the plates moving horizontally with speed 3×10^7 m/s. Axes shown: x to the right (i), y up (j), z out of page/toward viewer (k). For (iv): plates B (upper) and A (lower), 0·05 m long, with electron entering along line e; possible paths after plates: a (straight, undeflected), b (curving upward steeply), c (curving downward), d (upward at a smaller angle, straight line). See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 55/5/1.

Hints

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

Hint 1

Field between plates is V/d, pointing from + plate to − plate; an electron is pushed the opposite way.

Hint 2

Horizontal motion is uniform (t = l/v); vertical motion starts from rest with constant acceleration (y = ½at²).

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.

(i) $E = \frac{V}{d} = \frac{200}{0.01} = 2 \times 1 0^{4}$ V/m, directed from the positive (upper) plate to the negative (lower) plate, i.e. along $- \hat{k}$ . $E = - (2 \times 1 0^{4} V/m) \hat{k}$ — option (D).
1 mark
(ii) Force on electron is opposite to $E$ , i.e. along $+ \hat{k}$ . $a = \frac{e E}{m} = \frac{1.6 \times 1 0 ^{- 19} \times 2 \times 1 0 ^{4}}{9.1 \times 1 0 ^{- 31}} \approx 3.5 \times 1 0^{15}$ m s⁻². $a = (3.5 \times 1 0^{15} m s^{- 2}) \hat{k}$ — option (B).
1 mark
(iii)(a) Horizontal velocity is unchanged: $t = \frac{l}{v} = \frac{0.05}{3 \times 1 0 ^{7}} = 1.67 \times 1 0^{- 9}$ s — option (C).
1 mark
(iii)(b) [Alternative to (iii)(a), 1 mark] $y = \frac{1}{2} a t^{2} = \frac{1}{2} \times 3.5 \times 1 0^{15} \times (1.67 \times 1 0^{- 9})^{2} \approx 4.9 \times 1 0^{- 3}$ m $= 4.9$ mm — option (B). (Just less than 5 mm, so the electron just leaves the plates.)
(iv) Inside the plates the electron has uniform velocity horizontally and uniform acceleration towards the positive (upper) plate, so it follows a parabola curving upward — path b, option (B).
1 mark

Answer: (i) D; (ii) B; (iii)(a) C; (iii)(b) B; (iv) B

Where the marks are

(i) E = 2 × 10⁴ V/m along −k̂, option D (1 mark)
(ii) a = 3.5 × 10¹⁵ m/s² along +k̂, option B (1 mark)
(iii)(a) t = 1.67 × 10⁻⁹ s, option C; or (iii)(b) y ≈ 4.9 mm, option B (1 mark)
(iv) Parabolic path bending towards positive plate, path b, option B (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 55/5/1: every question, by section.

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CBSE Class 12 Physics 2026: question 29

Hints

Step-by-step solution

Revise the chapter

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