Match List I with List II
| LIST I | LIST II | ||
|---|---|---|---|
| A. | Electric potential V for a point charge | I. | Constant |
| B. | Electric potential (V) for an electric dipole on axial line | II. | \( V\propto\displaystyle\frac{1}{r^2} \) |
| C. | Electric potential inside a charged spherical conductor of radius r | III. | Zero |
| D. | Electric potential (V) on equatorial line of dipole | IV. | \( V\propto\displaystyle\frac{1}{r} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-II, C-I, D-III
(A) Electric potential V for a point charge — this matches (IV) \( V\propto\frac{1}{r} \). The potential of a point charge is \( V = \frac{q}{4\pi\varepsilon_0r} \).
(B) Electric potential (V) for an electric dipole on axial line — this matches (II) \( V\propto\frac{1}{r^{2}} \). For a short dipole on its axis \( V = \frac{p}{4\pi\varepsilon_0r^{2}} \), because the two opposite charges nearly cancel.
(C) Electric potential inside a charged spherical conductor of radius r — this matches (I) Constant. The field inside is zero, so no work is done in moving a charge there and the potential has the same value everywhere inside, equal to its value at the surface.
(D) Electric potential (V) on equatorial line of dipole — this matches (III) Zero. Every point on the equatorial line is equidistant from the two equal and opposite charges, so their potentials cancel.
NCERT (2026-27):
Potential Due to an Electric Dipole – “… The potential in the equatorial plane (θ = π/2) is zero. … (ii) The electric dipole potential falls off, at large distance, as 1/r2, not as 1/r, characteristic of the potential due to a single charge.”
Potential Due to a System of Charges – “… The electric field inside the shell is zero. This implies (Section 2.6) that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface
\( \displaystyle V = \frac{1}{4\pi\varepsilon_0}\frac{q}{R} \qquad [2.19(b)] \)”