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Q.What is the dipole moment of an electric dipole? What is the net charge on an electric dipole? State the SI unit of dipole moment of an electric dipole. Derive an expression for the electric field intensity at a point on the equatorial line of an electric dipole. (1+1+3=5) OR Two point charges −q-q and +q+q separated by a short distance 2a2a are placed in free space at points AA and BB respectively. Derive an expression for the electric potential at a point PP whose distance from the centre OO of the line ABAB is rr and OPOP makes an angle θ\theta with the electric dipole moment p⃗\vec p. Hence find the potential if PP lies on

(a) the axial line and
(b) the equatorial line. (3+1+1=5)
Meghalaya MboseMBOSE Meghalaya Intermediate Board 2025Subjective· 5mImportance★★★★★
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The dipole moment is p=q×2ap=q\times 2a, a vector from the negative to the positive charge, with SI unit C⋅\cdotm, though the net charge of a dipole is always zero; deriving the field at an equatorial point by adding the components of the fields due to both charges gives E=14πε0p(r2+a2)3/2E=\frac{1}{4\pi\varepsilon_0}\frac{p}{(r^2+a^2)^{3/2}}, directed opposite to p⃗\vec p. For the alternative, deriving potential due to a short dipole at general angle θ\theta and specializing gives kp/r2kp/r^2 on the axis and 0 on the equator.

Dipole moment, net charge, and SI unit

An electric dipole consists of two equal and opposite point charges, +q+q and −q-q, separated by a small distance 2a2a. Its dipole moment is a vector

p⃗=q(2a⃗)\vec p = q(2\vec a)

directed from the negative charge to the positive charge, with magnitude p=q(2a)p=q(2a).

The net charge on an electric dipole is (+q)+(−q)=0(+q)+(-q)=0 -- a dipole is electrically neutral overall.

The SI unit of dipole moment is the coulomb-metre (C⋅\cdotm).

Field at a point on the equatorial line

Let the dipole have charges −q-q at AA and +q+q at BB, separated by 2a2a, with centre OO. Let PP be a point on the equatorial line (perpendicular bisector of ABAB) at distance rr from OO.

Distance of PP from each charge:

AP=BP=r2+a2AP = BP = \sqrt{r^2+a^2}

Magnitude of field due to each charge at PP:

EA=EB=14πε0qr2+a2E_A = E_B = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2+a^2}

EAE_A (due to −q-q) points from PP towards AA; EBE_B (due to +q+q) points from BB towards PP (away from BB). By symmetry, the components of EAE_A and EBE_B perpendicular to the dipole axis (i.e., along POPO extended) cancel each other, while the components parallel to the dipole axis (both pointing in the direction from +q+q side to −q-q side, i.e. antiparallel to p⃗\vec p) add up.

Each field's component along the axis has magnitude EAcos⁡αE_A\cos\alpha, where cos⁡α=ar2+a2\cos\alpha = \dfrac{a}{\sqrt{r^2+a^2}} (with α\alpha the angle each of APAP, BPBP makes with the axis).

Eeq=2EAcos⁡α=2⋅14πε0qr2+a2⋅ar2+a2=14πε02qa(r2+a2)3/2E_{eq} = 2E_A\cos\alpha = 2\cdot\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2+a^2}\cdot\frac{a}{\sqrt{r^2+a^2}} = \frac{1}{4\pi\varepsilon_0}\frac{2qa}{(r^2+a^2)^{3/2}}

Since p=2qap=2qa:

Eeq=14πε0p(r2+a2)3/2E_{eq} = \frac{1}{4\pi\varepsilon_0}\frac{p}{(r^2+a^2)^{3/2}}

directed antiparallel to p⃗\vec p (i.e., from +q+q side towards −q-q side).

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