CBSE 2026 · Region 1 · Set 1 · Q31 · 5 marks
(a)An electric dipole consists of two point charges q and -q separated by a distance 2a. Derive an expression for the electric field $\displaystyle \overrightarrow{\mathrm{E}}$ due to this dipole at a point distant r from the centre of the dipole on the equatorial plane. Write the expression for the electric field at a far off point, i.e. $\displaystyle \mathrm{r} \gg \mathrm{a}$.A dipole is placed in x-y plane such that charges q and -q are located at $\displaystyle x=\mathrm{a}$ and $\displaystyle x=\mathrm{b}$ respectively. There exists an electric field $\displaystyle \overrightarrow{\mathrm{E}}=2 \hat{\mathrm{i}} \frac{\mathrm{N}}{\mathrm{C}}$ in the region. Calculate the force $\displaystyle \overrightarrow{\mathrm{F}}$ and torque $\displaystyle \vec{\tau}$ experienced by the dipole.Two cells of emf $\displaystyle \mathrm{E}_{1}$ and $\displaystyle \mathrm{E}_{2}$ with internal resistances $\displaystyle \mathrm{r}_{1}$ and $\displaystyle \mathrm{r}_{2}$ respectively, are connected in parallel by connecting their positive terminals together and negative terminals together. Deduce an expression for equivalent emf and equivalent internal resistance of the combination.(b)A parallel combination, as stated in (a) above, of two cells of emfs E and 3E and internal resistances R each is connected across a resistance 2R. Find the current that flows through resistance 2R.
(a)
An electric dipole consists of two point charges q and -q separated by a distance 2a. Derive an expression for the electric field $\displaystyle \overrightarrow{\mathrm{E}}$ due to this dipole at a point distant r from the centre of the dipole on the equatorial plane. Write the expression for the electric field at a far off point, i.e. $\displaystyle \mathrm{r} \gg \mathrm{a}$.
A dipole is placed in x-y plane such that charges q and -q are located at $\displaystyle x=\mathrm{a}$ and $\displaystyle x=\mathrm{b}$ respectively. There exists an electric field $\displaystyle \overrightarrow{\mathrm{E}}=2 \hat{\mathrm{i}} \frac{\mathrm{N}}{\mathrm{C}}$ in the region. Calculate the force $\displaystyle \overrightarrow{\mathrm{F}}$ and torque $\displaystyle \vec{\tau}$ experienced by the dipole.
Two cells of emf $\displaystyle \mathrm{E}_{1}$ and $\displaystyle \mathrm{E}_{2}$ with internal resistances $\displaystyle \mathrm{r}_{1}$ and $\displaystyle \mathrm{r}_{2}$ respectively, are connected in parallel by connecting their positive terminals together and negative terminals together. Deduce an expression for equivalent emf and equivalent internal resistance of the combination.
(b)
A parallel combination, as stated in (a) above, of two cells of emfs E and 3E and internal resistances R each is connected across a resistance 2R. Find the current that flows through resistance 2R.
Marking-scheme solution
(a)
The magnitude of the electric fields due to the two charges $\displaystyle +q$ and $\displaystyle -q$ are given by
$\displaystyle \mathrm{E}_{+\mathrm{q}}=\dfrac{\mathrm{q}}{4 \pi \varepsilon_{0}} \dfrac{1}{\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)}$
$\displaystyle \mathrm{E}_{-\mathrm{q}}=\dfrac{\mathrm{q}}{4 \pi \varepsilon_{0}} \dfrac{1}{\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)}$
The components normal to the dipole axis cancel away. The components along the dipole axis add up. The Total electric field is opposite to $\displaystyle \vec{p}$.
$\displaystyle \vec{\mathrm{E}}=-\left(\mathrm{E}_{+\mathrm{q}}+\mathrm{E}_{-\mathrm{q}}\right) \cos \theta(\hat{p})$
$\displaystyle \vec{\mathrm{E}}=-\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{2 \mathrm{qa}}{\left(\mathrm{r}^{2}+\mathrm{a}^{2}\right)^{3 / 2}} \hat{p}$
At large distance $\displaystyle \mathrm{r} \gg \mathrm{a}$
$\displaystyle \vec{\mathrm{E}}=\dfrac{-2 \mathrm{qa}}{4 \pi \varepsilon_{0} \mathrm{r}^{3}} \hat{p}$
$\displaystyle \because \vec{F}=\vec{F}_{+q}+\vec{F}_{-q}$
Net force $\displaystyle =\left[+q \cdot 2 \hat{i}-q \cdot 2 \hat{i}\right]$
$\displaystyle =0\ \mathrm{N}$
Torque $\displaystyle \vec{\tau}=\vec{p} \times \vec{\mathrm{E}}$
$\displaystyle \vec{\tau}=p(-\hat{i}) \times 2 \hat{i}$
$\displaystyle \tau=0$
Alternatively:
$\displaystyle \tau=p \mathrm{E} \sin \theta$
Angle between $\displaystyle \vec{p}$ and $\displaystyle \vec{\mathrm{E}}$ is $\displaystyle \pi$
$\displaystyle \tau=0$
Let $\displaystyle \mathrm{I}_{1}$ and $\displaystyle \mathrm{I}_{2}$ are the currents leaving from the positive electrodes of the cells $\displaystyle \varepsilon_{1}$ and $\displaystyle \varepsilon_{2}$ respectively, Hence $\displaystyle \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$
Potential difference across the terminals of cell $\displaystyle \varepsilon_{1}$ is
$\displaystyle \mathrm{V}=\varepsilon_{1}-\mathrm{I}_{1} \mathrm{r}_{1}$
Potential difference across the terminals of cell $\displaystyle \varepsilon_{2}$ is
$\displaystyle \mathrm{V}=\varepsilon_{2}-\mathrm{I}_{2} \mathrm{r}_{2}$
$\displaystyle \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$
$\displaystyle \mathrm{I}=\dfrac{\varepsilon_{1}-\mathrm{V}}{\mathrm{r}_{1}}+\dfrac{\varepsilon_{2}-\mathrm{V}}{\mathrm{r}_{2}}$
$\displaystyle \mathrm{I}=\left(\dfrac{\varepsilon_{1}}{\mathrm{r}_{1}}+\dfrac{\varepsilon_{2}}{\mathrm{r}_{2}}\right)-\mathrm{V}\left(\dfrac{1}{\mathrm{r}_{1}}+\dfrac{1}{\mathrm{r}_{2}}\right)$
$\displaystyle \mathrm{V}=\dfrac{\varepsilon_{1} \mathrm{r}_{2}+\varepsilon_{2} \mathrm{r}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}}-\mathrm{I}\left(\dfrac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$
$\displaystyle \mathrm{V}=\varepsilon_{\mathrm{eq}}-\mathrm{I} \mathrm{r}_{\mathrm{eq}}$
$\displaystyle \varepsilon_{\mathrm{eq}}=\left(\dfrac{\varepsilon_{1} \mathrm{r}_{2}+\varepsilon_{2} \mathrm{r}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$
$\displaystyle \mathrm{r}_{\mathrm{eq}}=\left(\dfrac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{1}+\mathrm{r}_{2}}\right)$
(b)
Equivalent emf
$\displaystyle \mathrm{E}_{\mathrm{eq}}=\dfrac{\varepsilon_{1} \mathrm{r}_{2}+\varepsilon_{2} \mathrm{r}_{1}}{\mathrm{r}_{1}+\mathrm{r}_{2}}$
$\displaystyle \mathrm{E}_{\mathrm{eq}}=\dfrac{\mathrm{E} \times \mathrm{R}+3 \mathrm{E} \times \mathrm{R}}{\mathrm{R}+\mathrm{R}}$
$\displaystyle \mathrm{E}_{\mathrm{eq}}=\dfrac{4 \mathrm{ER}}{2 \mathrm{R}}$
$\displaystyle \mathrm{E}_{\mathrm{eq}}=2 \mathrm{E}$
Equivalent resistance
$\displaystyle \mathrm{r}_{\mathrm{eq}}=\dfrac{\mathrm{r}_{1} \mathrm{r}_{2}}{\mathrm{r}_{2}+\mathrm{r}_{2}}$
$\displaystyle \mathrm{r}_{\mathrm{eq}}=\dfrac{\mathrm{RR}}{\mathrm{R}+\mathrm{R}}$
$\displaystyle \mathrm{r}_{\mathrm{eq}}=\dfrac{\mathrm{R}}{2}$
$\displaystyle \mathrm{I}=\dfrac{\mathrm{E}_{\mathrm{eq}}}{2 \mathrm{R}+\dfrac{\mathrm{R}}{2}}$
$\displaystyle \mathrm{I}=\dfrac{4 \mathrm{E}}{5 \mathrm{R}}\ \mathrm{A}$
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