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

Hints

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

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.

1. Gauss's theorem: the total electric flux through any closed surface equals $\frac{1}{{\epsilon }_0}$ times the net charge enclosed by it: $\oint \vec{E}\cdot d\vec{S}=\frac{q_{enc}}{{\epsilon }_0}$

   1 mark

2. Take an infinitely long straight wire with uniform linear charge density λ. To find E at a point P at distance r, choose a cylindrical Gaussian surface of radius r and length l, coaxial with the wire. By symmetry, E is radial and has the same magnitude at all points of the curved surface.

   ½ mark

3. $\oint \vec{E}\cdot d\vec{S}={\int }_{\text{curved}}EdS\cos 0^{\circ }+{\int }_{\text{two end faces}}EdS\cos 90^{\circ }$

   ½ mark

4. The flat ends contribute zero [E is parallel to the end faces], so $\varphi =E\times 2\pi rl$

   ½ mark

5. Charge enclosed $=\lambda l$. By Gauss's theorem $E\cdot 2\pi rl=\frac{\lambda l}{{\epsilon }_0}\Rightarrow E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$, directed radially outward for λ > 0.

   ½ mark

Answer: $\oint \vec{E}\cdot d\vec{S}=q/{\epsilon }_0$; $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$ (radial)

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