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Physics · 2025 · 5 marks
CBSE 2025 · Region 4 · Set 2 · Q33
(a) (i) Two point charges +q and -q are held at (a, 0) and ( - a, 0 ) in x - y plane. Obtain an expression for the net electric field due to the…
(i)
Two point charges +q and -q are held at (a, $\displaystyle 0$) and ( $\displaystyle -\mathrm{a}, 0$ ) in $\displaystyle \mathrm{x}-\mathrm{y}$ plane. Obtain an expression for the net electric field due to the charges at a point ( $\displaystyle 0, \mathrm{y}$ ) . Hence, find electric field at a far off point ( $\displaystyle \mathrm{y} \gg \mathrm{a}$ ) .
(ii)
Three point charges of $\displaystyle -2 \mathrm{nC},-1 \mathrm{nC}$, and +$\displaystyle 5$ nC are kept at the vertices $\displaystyle \mathrm{A}, \mathrm{B}$ and C of an equilateral triangle of side $\displaystyle 0.2$ m. Find the total amount of work done in shifting the charges from A to $\displaystyle \mathrm{A}_{1}, \mathrm{~B}$ to $\displaystyle \mathrm{B}_{1}$ and C to $\displaystyle \mathrm{C}_{1}$. Here $\displaystyle \mathrm{A}_{1}, \mathrm{~B}_{1}$ and $\displaystyle \mathrm{C}_{1}$ are the midpoints of sides $\displaystyle \mathrm{AB}, \mathrm{BC}$ and CA, respectively.
(i)
Show that Gauss's theorem is consistent with Coulomb's law. Using it, derive an expression for the electric field due to a uniformly charged thin spherical shell of radius $\displaystyle \mathrm{r}$ at a point at a distance $\displaystyle \mathrm{y}$ from the centre of the shell such that
(I)
$\displaystyle \mathrm{y}>\mathrm{r}$, and (II) $\displaystyle \mathrm{y}<\mathrm{r}$.
(ii)
A point charge of +$\displaystyle 2$ nC is kept at the origin of a three-dimensional coordinate system. Find the type and magnitude of the charge which should be kept at ( $\displaystyle 0,0,-6 \mathrm{~m}$ ) so that the potential due to the system becomes zero at $\displaystyle (0,0,2 \mathrm{~m})$.
Marking-scheme solution
(a)
Magnitude of electric field due to the two charges +q and −q are given by:
$\displaystyle E_{+q}=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{y^{2}+a^{2}}$
$\displaystyle E_{-q}=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{y^{2}+a^{2}}$
Components normal to the dipole axis cancel out.
The components along the dipole axis add up.
The total electric field is opposite to the dipole moment:
$\displaystyle \vec{E}=-\left(E_{+q}+E_{-q}\right) \cos \theta \hat{p}$
$\displaystyle =-\frac{2 q a}{4 \pi \varepsilon_{0}\left(y^{2}+a^{2}\right)^{3 / 2}} \hat{p}$ ($\displaystyle \hat{p}$ is a unit vector along the dipole moment)
At large distance $\displaystyle (y \gg a)$:
$\displaystyle \vec{E}=\frac{-2 q a}{4 \pi \varepsilon_{0} y^{3}} \hat{p}$
(ii)
Initial electrostatic potential energy of the system:
$\displaystyle U_{1}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{A} q_{B}}{A B}+\frac{q_{C} q_{A}}{A C}+\frac{q_{C} q_{B}}{B C}\right)$
$\displaystyle =\frac{9 \times 10^{9}}{0.2}[(-2 \times-1)+(-2 \times 5)+(-1 \times 5)] \times 10^{-18}$
$\displaystyle U_{1}=-5.85 \times 10^{-7} \mathrm{~J}$
$\displaystyle U_{2}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{q_{A_{1}} q_{B_{1}}}{A_{1} B_{1}}+\frac{q_{C_{1}} q_{A_{1}}}{A_{1} C_{1}}+\frac{q_{C_{1}} q_{B_{1}}}{B_{1} C_{1}}\right)$
$\displaystyle U_{2}=-11.7 \times 10^{-7} \mathrm{~J}$
$\displaystyle W=U_{2}-U_{1}=-5.85 \times 10^{-7} \mathrm{~J}$
(b)
Gauss's theorem is based on the inverse square dependence on distance contained in Coulomb's law.
According to Gauss's theorem:
$\displaystyle \oint \vec{E} \cdot d \vec{s}=\frac{q}{\varepsilon_{0}}$
$\displaystyle E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}$
According to Coulomb's law, force on charge $\displaystyle q_{0}$ in this field:
$\displaystyle F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q q_{0}}{r^{2}}$
Therefore, Gauss's law is consistent with Coulomb's law.
(I)
For $\displaystyle y>r$:
Electric flux through Gaussian surface $\displaystyle =E \times 4 \pi y^{2}$
The charge enclosed by the surface $\displaystyle =\sigma \times 4 \pi r^{2}$
Using Gauss theorem:
$\displaystyle E\left(4 \pi y^{2}\right)=\frac{\sigma 4 \pi r^{2}}{\varepsilon_{0}}$
$\displaystyle \vec{E}=\frac{q}{4 \pi \varepsilon_{0} y^{2}} \hat{r}$
(II)
For $\displaystyle y<r$:
The charge enclosed by the Gaussian surface $\displaystyle =0$
Using Gauss theorem:
Electric flux $\displaystyle =E\left(4 \pi y^{2}\right)=0$
i.e. $\displaystyle E=0 \quad(y<r)$
(ii)
Let the charge kept at A be q.
Potential at point B due to charge at the origin O and charge (q) at A:
$\displaystyle V=V_{1}+V_{2}$
$\displaystyle V=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{2 \times 10^{-9}}{2}+\frac{q}{6+2}\right]$
$\displaystyle \frac{1}{4 \pi \varepsilon_{0}}\left[10^{-9}+\frac{q}{8}\right]=0$
$\displaystyle q=-8 \times 10^{-9} \mathrm{C}$
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