Read the following statements about the electric potential due to an electric dipole of dipole moment \(\vec p\). Here \(\theta\) is the angle between the position vector \(\vec r\) and \(\vec p\).
(A) The result \(V=p\cos\theta/4\pi\varepsilon_0r^{2}\) is exact for a point dipole.
(B) On the dipole axis, the potential is positive on the side of the negative charge.
(C) The potential is zero at every point of the equatorial plane.
(D) The potential changes if the position vector is rotated about \(\vec p\) with \(\theta\) kept fixed.
Which of the above statements are incorrect? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (D) only
Far from a dipole, \( V=p\cos\theta/4\pi\varepsilon_0r^{2} \). The dipole moment points from \( -q \) to \( +q \).
(A) The result \( V=p\cos\theta/4\pi\varepsilon_0r^{2} \) is exact for a point dipole. — True. For a dipole of finite size the result holds only far away. For a point dipole it holds at every distance.
(B) On the dipole axis, the potential is positive on the side of the negative charge. — False. The side of \( +q \) is \( \theta=0 \), where \( V=+p/4\pi\varepsilon_0r^{2} \). On the side of \( -q \), \( \theta=\pi \) and \( \cos\theta=-1 \), so the potential is negative.
(C) The potential is zero at every point of the equatorial plane. — True. In the equatorial plane \( \theta=90^\circ \) and \( \cos90^\circ=0 \), so \( V=0 \).
(D) The potential changes if the position vector is rotated about \( \vec p \) with \( \theta \) kept fixed. — False. The potential depends only on \( r \) and \( \theta \). Such a rotation keeps both, so the potential stays the same. The potential is axially symmetric about \( \vec p \).
NCERT (2026-27):
Potential Due to an Electric Dipole – “The electric potential of a dipole is then given by”
\( \displaystyle V=\frac{1}{4\pi\varepsilon_0}\frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^{2}};\quad (r\gg a) \qquad (2.15) \)
“… For a point dipole p at the origin, Eq. (2.15) is, however, exact.”
Potential Due to an Electric Dipole – “… (Positive sign for θ = 0, negative sign for θ = π.) …”
Potential Due to an Electric Dipole – “… The potential in the equatorial plane (θ = π/2) is zero.”
Potential Due to an Electric Dipole – “… (It is, however, axially symmetric about p. That is, if you rotate the position vector r about p, keeping θ fixed, the points corresponding to P on the cone so generated will have the same potential as at P.)”