Step-by-step Derivations

To find the resultant electric field at point X, we will use the principle of superposition. Let's set up a coordinate system where the midpoint X is at the origin (0,0). The line joining the two dipoles is along the x-axis. The left dipole (Dipole 1) is at $(-r, 0)$ and the right dipole (Dipole 2) is at $(r, 0)$. The distance from each dipole to point X is $r$.

We will use the standard formulas for the electric field of an ideal dipole:

1. Axial Line: The electric field at a distance $d$ from the center of a dipole along its axis is given by $\vec{E}_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2\vec{p}}{d^3}$. The direction of the field is the same as the dipole moment vector $\vec{p}$.
2. Equatorial Line: The electric field at a distance $d$ from the center of a dipole along the line perpendicular to its axis (equatorial line) is given by $\vec{E}_{equatorial} = \frac{1}{4\pi\epsilon_0} \frac{-\vec{p}}{d^3}$. The direction of the field is opposite to the dipole moment vector $\vec{p}$.

Let's analyze each configuration from List-I:

Configuration (P):

1. Dipole 1 (left): The dipole moment is $\vec{p}_1 = p\hat{\jmath}$. Point X lies on the equatorial line of this dipole. The field at X due to Dipole 1 is: $\vec{E}_1 = \frac{-\vec{p}_1}{4\pi\epsilon_0 r^3} = -\frac{p}{4\pi\epsilon_0 r^3}\hat{\jmath}$
2. Dipole 2 (right): The dipole moment is $\vec{p}_2 = p\hat{\jmath}$. Point X also lies on the equatorial line of this dipole. The field at X due to Dipole 2 is: $\vec{E}_2 = \frac{-\vec{p}_2}{4\pi\epsilon_0 r^3} = -\frac{p}{4\pi\epsilon_0 r^3}\hat{\jmath}$
3. Resultant Field: The total electric field at X is the vector sum $\vec{E}_P = \vec{E}_1 + \vec{E}_2$. $\vec{E}_P = \left(-\frac{p}{4\pi\epsilon_0 r^3}\hat{\jmath}\right) + \left(-\frac{p}{4\pi\epsilon_0 r^3}\hat{\jmath}\right) = -\frac{2p}{4\pi\epsilon_0 r^3}\hat{\jmath} = -\frac{p}{2\pi\epsilon_0 r^3}\hat{\jmath}$ This matches option (2) in List-II. Thus, $\mathbf{P \rightarrow 2}$.

Configuration (Q):

1. Dipole 1 (left): The dipole moment is $\vec{p}_1 = p\hat{\jmath}$. Point X is on its equatorial line. $\vec{E}_1 = \frac{-\vec{p}_1}{4\pi\epsilon_0 r^3} = -\frac{p}{4\pi\epsilon_0 r^3}\hat{\jmath}$
2. Dipole 2 (right): The dipole moment is $\vec{p}_2 = -p\hat{\jmath}$. Point X is on its equatorial line. $\vec{E}_2 = \frac{-\vec{p}_2}{4\pi\epsilon_0 r^3} = -\frac{(-p\hat{\jmath})}{4\pi\epsilon_0 r^3} = \frac{p}{4\pi\epsilon_0 r^3}\hat{\jmath}$
3. Resultant Field: $\vec{E}_Q = \vec{E}_1 + \vec{E}_2 = \left(-\frac{p}{4\pi\epsilon_0 r^3}\hat{\jmath}\right) + \left(\frac{p}{4\pi\epsilon_0 r^3}\hat{\jmath}\right) = 0$ This matches option (1) in List-II. Thus, $\mathbf{Q \rightarrow 1}$.

Configuration (R):

1. Dipole 1 (left): The dipole moment is $\vec{p}_1 = p\hat{\imath}$. Point X lies on the axial line of this dipole. $\vec{E}_1 = \frac{2\vec{p}_1}{4\pi\epsilon_0 r^3} = \frac{2p}{4\pi\epsilon_0 r^3}\hat{\imath}$
2. Dipole 2 (right): The dipole moment shown in the diagram is $\vec{p}_2 = -p\hat{\jmath}$. Point X is on its equatorial line. $\vec{E}_2 = \frac{-\vec{p}_2}{4\pi\epsilon_0 r^3} = -\frac{(-p\hat{\jmath})}{4\pi\epsilon_0 r^3} = \frac{p}{4\pi\epsilon_0 r^3}\hat{\jmath}$
3. Resultant Field (based on diagram): $\vec{E}_R = \vec{E}_1 + \vec{E}_2 = \frac{p}{4\pi\epsilon_0 r^3}(2\hat{\imath} + \hat{\jmath})$ This result does not match any option in List-II. However, option (4) is $\vec{E} = \frac{p}{4\pi\epsilon_0\,r^3}\,(2\hat{\imath} - \hat{\jmath})$. This would be the result if the dipole moment of the right dipole was $\vec{p}_2 = p\hat{\jmath}$ (pointing up). Assuming a typo in the diagram for (R), where the right dipole should point up: $\vec{E}_2' = \frac{-(p\hat{\jmath})}{4\pi\epsilon_0 r^3} = -\frac{p}{4\pi\epsilon_0 r^3}\hat{\jmath}$ $\vec{E}_R' = \vec{E}_1 + \vec{E}_2' = \frac{p}{4\pi\epsilon_0 r^3}(2\hat{\imath} - \hat{\jmath})$ This matches option (4) in List-II. Thus, assuming a typo, $\mathbf{R \rightarrow 4}$.

Configuration (S):

1. Dipole 1 (left): The dipole moment is $\vec{p}_1 = p\hat{\imath}$. Point X is on its axial line. $\vec{E}_1 = \frac{2\vec{p}_1}{4\pi\epsilon_0 r^3} = \frac{2p}{4\pi\epsilon_0 r^3}\hat{\imath}$
2. Dipole 2 (right): The dipole moment shown in the diagram is $\vec{p}_2 = -p\hat{\imath}$. Point X is on its axial line. $\vec{E}_2 = \frac{2\vec{p}_2}{4\pi\epsilon_0 r^3} = \frac{2(-p\hat{\imath})}{4\pi\epsilon_0 r^3} = -\frac{2p}{4\pi\epsilon_0 r^3}\hat{\imath}$
3. Resultant Field (based on diagram): $\vec{E}_S = \vec{E}_1 + \vec{E}_2 = \left(\frac{2p}{4\pi\epsilon_0 r^3}\hat{\imath}\right) + \left(-\frac{2p}{4\pi\epsilon_0 r^3}\hat{\imath}\right) = 0$ This would match option (1), which is already matched with Q. Let's check option (5): $\vec{E} = \frac{p}{\pi\epsilon_0\,r^3}\,\hat{\imath} = \frac{4p}{4\pi\epsilon_0 r^3}\hat{\imath}$. This result would be obtained if both dipoles were pointing to the right, i.e., $\vec{p}_2 = p\hat{\imath}$. Assuming a typo in the diagram for (S), where the right dipole should also point right: $\vec{E}_2' = \frac{2(p\hat{\imath})}{4\pi\epsilon_0 r^3} = \frac{2p}{4\pi\epsilon_0 r^3}\hat{\imath}$ $\vec{E}_S' = \vec{E}_1 + \vec{E}_2' = \frac{2p\hat{\imath}}{4\pi\epsilon_0 r^3} + \frac{2p\hat{\imath}}{4\pi\epsilon_0 r^3} = \frac{4p}{4\pi\epsilon_0 r^3}\hat{\imath} = \frac{p}{\pi\epsilon_0 r^3}\hat{\imath}$ This matches option (5) in List-II. Thus, assuming a typo, $\mathbf{S \rightarrow 5}$.

Conclusion

Based on our analysis, correcting for likely typos in the diagrams for (R) and (S):

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

This set of matches corresponds to option C.