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Physics · 2026 · 5 marks
CBSE 2026 · Region 1 · Set 1 · Q31
(a) An electric dipole consists of two point charges q and -q separated by a distance 2a. Derive an expression for the electric field E due to this…
(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. $\displaystyle \mathrm{r} \gg \mathrm{a}$.
(b)
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.
(a)
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 $\displaystyle \mathrm{r} \gg \mathrm{a}$
$\displaystyle \vec{\mathrm{E}}=\dfrac{-2 \mathrm{qa}}{4 \pi \varepsilon_{0} \mathrm{r}^{3}} \hat{p}$
(b)
$\displaystyle \because \vec{F}=\vec{F}_{+q}+\vec{F}_{-q}$
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 $\displaystyle \pi$
$\displaystyle \tau=0$
(a)
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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