CBSE 2024 · Region 5 · Set 1 · Q31 · 5 marks
(i)Draw equipotential surfaces for an electric dipole.(ii)Two point charges $\displaystyle q_{1}$ and $\displaystyle q_{2}$ are located at $\displaystyle \vec{r}_{1}$ and $\displaystyle \vec{r}_{2}$ respectively in an external electric field $\displaystyle \vec{\mathrm{E}}$. Obtain an expression for the potential energy of the system.(iii)The dipole moment of a molecule is $\displaystyle 10^{-30} \mathrm{Cm}$. It is placed in an electric field $\displaystyle \overrightarrow{\mathrm{E}}$ of $\displaystyle 10^{5} \mathrm{~V} / \mathrm{m}$ such that its axis is along the electric field. The direction of $\displaystyle \vec{\mathrm{E}}$ is suddenly changed by $\displaystyle 60^{\circ}$ at an instant. Find the change in the potential energy of the dipole, at that instant.OR 31. (b) (i) A thin spherical shell of radius R has a uniform surface charge density $\displaystyle \sigma$. Using Gauss' law, deduce an expression for electric field (i) outside and (ii) inside the shell.(ii)Two long straight thin wires AB and CD have linear charge densities $\displaystyle 10 \mu \mathrm{C} / \mathrm{m}$ and $\displaystyle -20 \mu \mathrm{C} / \mathrm{m}$, respectively. They are kept parallel to each other at a distance $\displaystyle 1$ m . Find magnitude and direction of the net electric field at a point midway between them.
(i)
Draw equipotential surfaces for an electric dipole.
(ii)
Two point charges $\displaystyle q_{1}$ and $\displaystyle q_{2}$ are located at $\displaystyle \vec{r}_{1}$ and $\displaystyle \vec{r}_{2}$ respectively in an external electric field $\displaystyle \vec{\mathrm{E}}$. Obtain an expression for the potential energy of the system.
(iii)
The dipole moment of a molecule is $\displaystyle 10^{-30} \mathrm{Cm}$. It is placed in an electric field $\displaystyle \overrightarrow{\mathrm{E}}$ of $\displaystyle 10^{5} \mathrm{~V} / \mathrm{m}$ such that its axis is along the electric field. The direction of $\displaystyle \vec{\mathrm{E}}$ is suddenly changed by $\displaystyle 60^{\circ}$ at an instant. Find the change in the potential energy of the dipole, at that instant.
OR 31. (b) (i) A thin spherical shell of radius R has a uniform surface charge density $\displaystyle \sigma$. Using Gauss' law, deduce an expression for electric field (i) outside and (ii) inside the shell.
(ii)
Two long straight thin wires AB and CD have linear charge densities $\displaystyle 10 \mu \mathrm{C} / \mathrm{m}$ and $\displaystyle -20 \mu \mathrm{C} / \mathrm{m}$, respectively. They are kept parallel to each other at a distance $\displaystyle 1$ m . Find magnitude and direction of the net electric field at a point midway between them.
Marking-scheme solution
(ii)
Work done in bringing a charge $\displaystyle q_{1}$ from infinity to $\displaystyle r_{1}$:
$\displaystyle W_{1}=q_{1} V\left(r_{1}\right)$ -------------($\displaystyle 1$)
Work done in bringing a charge $\displaystyle q_{2}$ from infinity to $\displaystyle r_{2}$ against the external field:
$\displaystyle W_{2}=q_{2} V\left(r_{2}\right)$ -------------($\displaystyle 2$)
Work done on $\displaystyle q_{2}$ against the field due to $\displaystyle q_{1}$:
$\displaystyle W_{12}=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r_{12}}$ -----------------($\displaystyle 3$)
Potential energy of the system = total work done
$\displaystyle =q_{1} V\left(r_{1}\right)+q_{2} V\left(r_{2}\right)+\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} r_{12}}$
(iii)
Change in potential energy = work done
$\displaystyle W=p E\left[\cos \theta_{0}-\cos \theta_{1}\right]$
$\displaystyle W=10^{-30} \times 10^{5}\left[\cos 0^{\circ}-\cos 60^{\circ}\right]$
$\displaystyle W=5.0 \times 10^{-26} \mathrm{~J}$
(i)
Electric field outside the shell:
Electric flux through Gaussian surface $\displaystyle \Phi=E \times 4 \pi r^{2}$
Charge enclosed by the Gaussian surface $\displaystyle Q=\sigma \times 4 \pi R^{2}$
Using Gauss' law: $\displaystyle \oint \vec{E} \cdot d \vec{s}=\frac{Q}{\varepsilon_{0}}$
$\displaystyle E \times 4 \pi r^{2}=\frac{\sigma 4 \pi R^{2}}{\varepsilon_{0}}$
$\displaystyle \therefore E=\frac{\sigma R^{2}}{\varepsilon_{0} r^{2}}$
$\displaystyle \vec{E}=\frac{\sigma R^{2}}{\varepsilon_{0} r^{2}} \hat{r}$
Field inside the shell:
Electric flux through Gaussian surface $\displaystyle \Phi=E \times 4 \pi r^{2}$ $\displaystyle (r<R)$
Charge enclosed by the Gaussian surface $\displaystyle Q=0$
By Gauss' law: $\displaystyle E \times 4 \pi r^{2}=0$
i.e. $\displaystyle E=0$
(ii)
Electric field due to a long straight charged wire of linear charge density $\displaystyle \lambda$:
$\displaystyle E=\frac{\lambda}{2 \pi \varepsilon_{0} r}$
Net electric field at the mid-point:
$\displaystyle E_{n e t}=E_{1}+E_{2}$
$\displaystyle =\frac{\lambda_{1}}{2 \pi \varepsilon_{0} r}+\frac{\lambda_{2}}{2 \pi \varepsilon_{0} r}$
$\displaystyle E_{n e t}=\frac{1}{2 \pi \varepsilon_{0} r}\left[\lambda_{1}+\lambda_{2}\right]$
$\displaystyle =\frac{2 \times 9 \times 10^{9}}{0.5}[10+20] \times 10^{-6}$
$\displaystyle =1.08 \times 10^{6} \mathrm{~NC}^{-1}$
$\displaystyle E_{n e t}$ is directed towards CD.
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.