CBSE Class 12 Physics 2025: question 31(b)

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Step-by-step solution

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1. (i)(I) Axial point P at distance x from the centre, on the side of +q. Distance from +q is $(x-a)$ and from −q is $(x+a)$. So $V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{x-a}-\frac{q}{x+a}\right]$.

   ½ mark

2. $V=\frac{1}{4\pi {\epsilon }_0}\frac{q(2a)}{x^2-a^2}=\frac{1}{4\pi {\epsilon }_0}\frac{p}{x^2-a^2}$, where $p=q\cdot 2a$.

   ½ mark

3. For $x\gg a$: $V=\frac{1}{4\pi {\epsilon }_0}\frac{p}{x^2}$.

   ½ mark

4. (II) Equatorial (bisector) point at distance x from the centre. Both charges are at the same distance $\sqrt{x^2+a^2}$.

   ½ mark

5. $V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q}{\sqrt{x^2+a^2}}-\frac{q}{\sqrt{x^2+a^2}}\right]$

   ½ mark

6. $V=0$ at every point on the bisector.

   ½ mark

7. (ii) $\vec{\tau }=\vec{p}\times \vec{E}=(0.8\hat{i}+0.6\hat{j})\times 10^{-29}\times (1.0\times 10^7\hat{k})$. Using $\hat{i}\times \hat{k}=-\hat{j}$ and $\hat{j}\times \hat{k}=\hat{i}$: $\vec{\tau }=(0.6\hat{i}-0.8\hat{j})\times 10^{-22}$ N m.

   ½ mark

8. $|\vec{\tau }|=\sqrt{0.6^2+0.8^2}\times 10^{-22}$

   ½ mark

9. $|\vec{\tau }|=1.0\times 10^{-22}$ N m.

   ½ mark

10. Angle of $\vec{\tau }$ with the x-axis: $\tan \theta =\frac{0.8}{0.6}=\frac{4}{3}\Rightarrow \theta \approx 53^{\circ }$, measured below the x-axis towards −y. (The dipole itself makes ${\tan }^{-1}(0.6/0.8)\approx 37^{\circ }$ with the x-axis.)

    ½ mark

Answer: (I) V = p/(4πε₀x²) on the axis; (II) V = 0 on the bisector; (ii) |τ| = 1.0 × 10⁻²² N m, at about 53° to the x-axis (towards −y).

Revise the chapter

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