CBSE Class 12 Physics 2026: question 22(a)

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Step-by-step solution

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1. (a) Take an infinite thin sheet with uniform surface charge density $\sigma$. By symmetry, $\vec{E}$ is normal to the sheet, points away from it on both sides (for $\sigma >0$), and has the same magnitude at equal distances. Choose as the Gaussian surface a cylinder of cross-section area $A$ whose axis is normal to the sheet and which extends equally on both sides.

   ½ mark

2. Flux through the curved surface is zero because $\vec{E}\perp d\vec{A}$ there. Flux through the two flat faces $=EA+EA=2EA$.

   ½ mark

3. 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, so $\vec{E}=\frac{\sigma }{2{\epsilon }_0}\hat{n}$.

   1 mark

4. (b)(i) Between the sheets the two fields, each $\frac{\sigma }{2{\epsilon }_0}$, are equal and opposite: $E_{inside}=\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}=0$.

   ½ mark

5. (b)(ii) Outside the sheets the two fields point the same way: $E_{outside}=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}$, directed away from the sheets.

   ½ mark

Answer: $E=\sigma /2{\epsilon }_0$; (i) zero inside; (ii) $\sigma /{\epsilon }_0$ outside

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