CBSE Class 12 Physics 2024: question 31(b)

Hints

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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. (i) **Outside ($r>R$):** take a concentric spherical Gaussian surface of radius $r$. By symmetry $\vec{E}$ is radial and has the same magnitude everywhere on it.

   Flux through the surface: $\varphi =E\cdot 4\pi r^2$.

   ½ mark

2. Charge enclosed: $Q=\sigma \cdot 4\pi R^2$.

   ½ mark

3. Gauss's law: $\oint \vec{E}\cdot d\vec{S}=\frac{Q}{{\epsilon }_0}$, so $E\cdot 4\pi r^2=\frac{\sigma 4\pi R^2}{{\epsilon }_0}$.

   ½ mark

4. $$\vec{E}=\frac{\sigma R^2}{{\epsilon }_0{r}^2}\hat{r}(r>R)$$

   ½ mark

5. **Inside ($r<R$):** take a Gaussian sphere of radius $r<R$. Flux $=E\cdot 4\pi r^2$.

   The charge enclosed is $Q=0$, since all the charge lies on the shell.

   ½ mark

6. So $E\cdot 4\pi r^2=0$, which gives $E=0$ everywhere inside the shell.

   ½ mark

7. (ii) Field of a long charged wire: $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$. The midpoint is at $r=0.5$ m from each wire.

   ½ mark

8. Directions at the midpoint:

   • The field of AB ($+10\mu$C/m) points away from AB, i.e. towards CD.
   • The field of CD ($-20\mu$C/m) points towards CD.

   Both are along the same direction, so they add:
   $E_{net}=\frac{1}{2\pi {\epsilon }_{0}r}({\lambda }_1+|{\lambda }_2|)$

   ½ mark

9. $E_{net}=\frac{2\times 9\times 10^9\times (10+20)\times 10^{-6}}{0.5}$

   ½ mark

10. $E_{net}=1.08\times 10^6{\text{N C}}^{-1}$, directed towards CD (perpendicular to the wires).

    ½ mark

Answer: (i) E = σR²/(ε₀r²) outside (radial), E = 0 inside; (ii) 1.08 × 10⁶ N/C directed towards CD

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