Question 137: Electrostats & Capacitance

Question

List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude $p$, oriented as marked by arrows in the figures. In all the configurations the dipoles are fixed such that they are at a distance $2 r$ apart along the $x$ direction. The midpoint of the line joining the two dipoles is $X$. The possible resultant electric fields $\vec{E}$ at $X$ are given in List-II.
Choose the option that describes the correct match between the entries in List-I to those in List-II.

List–I
List–II

(P) 
(1) $$
\vec{E}=0
$$

(Q) 
(2) $\displaystyle \vec{E} = -\,\frac{p}{2\pi\epsilon_0\,r^3}\,\hat{\jmath}$

(R) 
(3) $\displaystyle \vec{E} = -\,\frac{p}{4\pi\epsilon_0\,r^3}\,(\hat{\imath} - \hat{\jmath})$

(S) 
(4) $\displaystyle \vec{E} = \frac{p}{4\pi\epsilon_0\,r^3}\,(2\hat{\imath} - \hat{\jmath})$

(5) $\displaystyle \vec{E} = \frac{p}{\pi\epsilon_0\,r^3}\,\hat{\imath}$

Accepted Answer

The electric field $\vec{E}$ due to a dipole at a point is given by:
On the axial line (along the dipole axis): $E_{	ext {axial }}=\frac{2 k p}{r^3}$ (along the dipole moment )
where, $K=\frac{1}{4 \pi \varepsilon_0}$
On the equatorial line (perpendicular bisector):
$$ 	ext { Equatorial }=\frac{-k p}{r^3}\binom{	ext { opposite to }}{	ext { the dipole moment }} $$
Here, $x$ is on the equatorial line for both dipoles:
$$ \vec{E}=\overrightarrow{E_1}+\overrightarrow{E_2} $$
$$ 	ext { } \begin{aligned} & So\,\,\, \vec{E}=-\frac{k p}{r^3} \hat{j}-\frac{k p}{r^3} \hat{j} \ & \Rightarrow \vec{E}=-\frac{2 k p}{r^3} \hat{j}=\frac{-2}{r^3}\left(\frac{1}{4 \pi \varepsilon_0}ight) p{\hat{j}} \ & \Rightarrow \vec{E}=\frac{-p}{2 \pi \varepsilon_0 r^3} \hat{j} \ & 	ext { So }(p) ightarrow(2) \end{aligned} $$
Here, $x$ is on the equatorial line for both dipoles.
$$ \begin{aligned} &	ext { Electric field at } x 	ext { due to dipole } 1 	ext {, }\ &\vec{E}_1=-\frac{k p}{r^3} \hat{j} \end{aligned} $$
Electric field at $x$ due to dipole 2,
$$ \vec{E}_2=\frac{k p}{r^3} \hat{j} \quad\left(\begin{array}{l} 	ext { as dipole moment } \ 	ext { is in }-\hat{j} 	ext { direction }) \end{array}ight. $$
so. $\vec{E}=\vec{E}_1+\vec{E}_2$
$$ \begin{aligned} & =\frac{-k p}{r^3} \hat{j}+\frac{k p}{r^3} \hat{j} \ & \Rightarrow \vec{E}=0 \end{aligned} $$
So $($ Q $) \longrightarrow(1)$
Here, $x$ is on the equatorial
line for dipole 1 and on the axial line for dipole 2.
So, $\vec{E}_1=\frac{-k p}{r^3} \hat{j}$
and $\vec{E}_2=\frac{2 k p}{r^3} \hat{i}$
$$ \begin{aligned}So\,\,\,\,\, \vec{E} & =\vec{E}_1+\vec{E}_2 \ & =\frac{-k p}{r^3} \hat{j}+\frac{2 k p}{r^3} \hat{i} \ & =\frac{k p}{r^3}(-\hat{j}+2 \hat{i}) \end{aligned} $$
$$ \begin{aligned} &\begin{aligned} & \Rightarrow \vec{E}=\frac{k p}{r^3}(2 \hat{i}-\hat{j}) \ & \Rightarrow \vec{E}=\frac{p}{4 \pi \varepsilon_0 r^3}(2 \hat{i}-\hat{j}) \end{aligned}\ &	ext { So }(R) ightarrow(4) \end{aligned} $$
Here, $x$ is on the axial line for both dipoles.
so $\vec{E}_1=\frac{2 k p}{r^3} \hat{i}$ and $\overrightarrow{E_2}=\frac{2 k p}{r^3} \hat{i}$
So
$$ \begin{aligned} \vec{E} & =\vec{E}_1+\vec{E}_2 \ & =\frac{2 k p}{r^3} \hat{i}+\frac{2 k p}{r^3} \hat{i} \ \Rightarrow \vec{E} & =\frac{4 k p}{r^3} \hat{i} \ \Rightarrow \vec{E} & =4\left(\frac{1}{4 \pi \varepsilon_0}ight) \frac{p}{r^3} \hat{i} \ \Rightarrow \vec{E} & =\frac{p}{\pi \varepsilon_0 r^3} \hat{i} \end{aligned} $$
So $(S) ightarrow(5)$
Hence, option (c) is correct.

Answer

$ \mathrm{P} \rightarrow 3, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 2, \mathrm{~S} \rightarrow 4$

Answer

$\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 5, \mathrm{R} \rightarrow 3, \mathrm{~S} \rightarrow 1$

Answer

$\mathrm{P} \rightarrow 2, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 4, \mathrm{~S} \rightarrow 5$

Answer

$\mathrm{P} \rightarrow 2, \mathrm{Q} \rightarrow 1, \mathrm{R} \rightarrow 3, \mathrm{~S} \rightarrow 5$

Electrostats & Capacitance

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