A cube of side 0.1 m is placed, as shown in the figure, in a region where electric field E = 500 x i exists. Here x is in meters and E in NC^-1. Calculate : (a) the flux passing through the cube, and (b) the charge within the cube.

PhysicsCBSE Class 12 · Board exam 2024, set 55/1/1 · 3 marks · Short answer

CBSE Class 12 Physics 2024: question 22

Set in the CBSE Class 12 board exam of 2024, paper 55/1/1, section C, question 22, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q22 · 3 marks · Short answerElectric Charges and Fields

A cube of side 0.1 m is placed, as shown in the figure, in a region where electric field $E = 500 x \hat{i}$ exists. Here x is in meters and E in $N C^{- 1}$ . Calculate :
(a) the flux passing through the cube, and
(b) the charge within the cube.

Figure in the paper: Axes x, y, z with origin O. A cube of side 0.1 m is placed with one face parallel to the y-z plane, at a distance 0.1 m from the origin along the x-axis. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 55/1/1.

Hints

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

Hint 1

The field points only along x, so only two faces of the cube matter.

Hint 2

Find E at x = 0.1 m and x = 0.2 m, and take the outward normal of each face into account.

Hint 3

Gauss's law links the net flux to the enclosed charge.

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.

(a) $E = 500 x \hat{i}$ is along x, so flux passes only through the two faces perpendicular to x. Each has area $A = (0.1)^{2} = 0.01 m^{2}$ . The left face is at $x = 0.1$ m and the right face at $x = 0.2$ m.
Left face (outward normal along $- \hat{i}$ ): $E_{L} = 500 \times 0.1 = 50$ N/C, so $ϕ_{L} = - 50 \times 0.01 = - 0.5 N m^{2} C^{- 1}$ .
½ mark
Right face (outward normal along $+ \hat{i}$ ): $E_{R} = 500 \times 0.2 = 100$ N/C, so $ϕ_{R} = 100 \times 0.01 = 1.0 N m^{2} C^{- 1}$ .
½ mark
The other four faces are parallel to $E$ and carry zero flux. Net flux: $ϕ = ϕ_{L} + ϕ_{R} = 0.5 N m^{2} C^{- 1}$ .
1 mark
(b) By Gauss's law, $ϕ = \frac{q}{ε _{0}}$ , so $q = ε_{0} ϕ = 8.85 \times 1 0^{- 12} \times 0.5$ .
½ mark
$q \approx 4.4 \times 1 0^{- 12}$ C
½ mark

Answer: (a) 0.5 N m² C⁻¹; (b) ≈ 4.4 × 10⁻¹² C

Where the marks are

Flux through the left face = −0.5 N m²/C (½ mark)
Flux through the right face = 1 N m²/C (½ mark)
Net flux = 0.5 N m²/C (1 mark)
Uses Gauss's law $q = ε_{0} ϕ$ (½ mark)
Charge ≈ 4.4 × 10⁻¹² C (½ mark)

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Chapter: Electric Charges and Fields (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Electric Charges and Fields, one at a time, marked against the answer key and marking points.Back to the full paper2024 board paper, set 55/1/1: every question, by section.

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CBSE Class 12 Physics 2024: question 22

Hints

Step-by-step solution

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