CBSE Class 12 Physics sample paper 2024-25: question 28(II)

Hints

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Step-by-step solution

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1. (a) Electric flux through a surface is the total number of electric field lines crossing it normally, i.e. the surface integral of the field: $\varphi =\int \vec{E}\cdot d\vec{S}$. It is a scalar. SI unit: N m² C⁻¹ (equivalently V m).

   1 mark

2. (b) Take an infinite plane sheet with uniform surface charge density σ. By symmetry, E is perpendicular to the sheet and has the same magnitude on both sides. Choose a Gaussian cylinder of cross-section A piercing the sheet, with its flat faces equidistant from it on either side and point P on one face.

   ½ mark

3. $\varphi ={\int }_{\text{RCF}}\vec{E}\cdot d\vec{A}+{\int }_{\text{LCF}}\vec{E}\cdot d\vec{A}+{\int }_{\text{CS}}\vec{E}\cdot d\vec{A}=\int EdA\cos 0^{\circ }+\int EdA\cos 0^{\circ }+\int EdA\cos 90^{\circ }$

   ½ mark

4. $\varphi =EA+EA+0=2EA$ [the curved surface is parallel to E]

   ½ mark

5. Charge enclosed $=\sigma A$. By Gauss's law $2EA=\frac{\sigma A}{{\epsilon }_0}\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}$, directed normally away from the sheet for σ > 0 and independent of distance.

   ½ mark

Answer: (a) $\varphi =\oint \vec{E}\cdot d\vec{S}$, unit N m² C⁻¹; (b) $E=\frac{\sigma }{2{\epsilon }_0}$

Revise the chapter

• Chapter: Electric Charges and Fields (notes, formulas, where marks are lost and every past question)
• Topic: Gauss's Law and Its Applications