CBSE Class 12 Physics 2026: question 31(a)

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Step-by-step solution

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1. (a) Place $-q$ at B and $+q$ at A, separated by $2a$, with centre O. Let P lie on the equatorial line with OP $=r$. Each charge is at distance $\sqrt{r^2+a^2}$ from P, so
   $|{\vec{E}}_{+q}|=|{\vec{E}}_{-q}|=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2+a^2}$

   1 mark

2. Resolve the two fields. Their components normal to the dipole axis cancel. The components along the axis add, both pointing from $+q$ towards $-q$, i.e. opposite to $\vec{p}$. If $\theta$ is the angle each field makes with the axis, $\cos \theta =\frac{a}{\sqrt{r^2+a^2}}$.

   ½ mark

3. $E=2E_{+q}\cos \theta =\frac{2qa}{4\pi {\epsilon }_0(r^2+a^2{)}^{3/2}}$. With $\vec{p}=q(2a)\hat{p}$: $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0(r^2+a^2{)}^{3/2}}$

   1 mark

4. For $r\gg a$: $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0{r}^3}$

   ½ mark

5. (b) The field $\vec{E}=2\hat{i}$ N/C is uniform. Net force: $\vec{F}=q\vec{E}+(-q)\vec{E}=0$.

   1 mark

6. Dipole moment (from $-q$ to $+q$): $\vec{p}=q(a-b)\hat{i}$. Torque: $\vec{\tau }=\vec{p}\times \vec{E}=q(a-b)\cdot 2(\hat{i}\times \hat{i})=0$, since $\vec{p}\parallel \vec{E}$.

   1 mark

Answer: $\vec{E}=-\frac{\vec{p}}{4\pi {\epsilon }_0(r^2+a^2{)}^{3/2}}$, which becomes $-\frac{\vec{p}}{4\pi {\epsilon }_0{r}^3}$ for $r\gg a$; (b) $\vec{F}=0$, $\vec{\tau }=0$

Revise the chapter

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