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Q.Derive an expression for the intensity of the electric field at a point on the equatorial plane of an electric dipole.

Andhra Pradesh BieapBIEAP Intermediate Board 2024Subjective· 4mImportance★★★★★
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On the equatorial line, the perpendicular field components cancel and the axial ones add, giving E=p4πε0(r2+a2)3/2E = \dfrac{p}{4\pi\varepsilon_0 (r^2+a^2)^{3/2}}, which reduces to p4πε0r3\dfrac{p}{4\pi\varepsilon_0 r^3} for r≫ar\gg a, directed opposite to the dipole moment.

Setup (NCERT/CBSE electric-charges-and-fields):

A dipole consists of charges +q+q and −q-q separated by 2a2a; its dipole moment is p=q(2a)p = q(2a), directed from −q-q to +q+q. Consider a point P on the equatorial plane — the perpendicular bisector of the dipole axis — at distance rr from the centre O.

Step 1 — distance from each charge to P:

Each charge is at distance

r′=r2+a2r' = \sqrt{r^2 + a^2}

from P.

Step 2 — magnitude of each field:

The fields due to +q+q and −q-q have equal magnitudes:

E+=E−=14πε0q(r2+a2)E_{+} = E_{-} = \frac{1}{4\pi\varepsilon_0}\frac{q}{(r^2+a^2)}

E+E_+ points away from +q+q, E−E_- points toward −q-q.

Step 3 — resolve into components:

Resolve each field into a component parallel to the dipole axis and a component perpendicular to it (along OP).

  • The perpendicular (along-OP) components are equal and opposite, so they cancel.
  • The components parallel to the axis (pointing from +q+q toward −q-q, i.e. anti-parallel to p⃗\vec{p}) add up. Each contributes a factor cos⁡θ\cos\theta, where cos⁡θ=ar2+a2\cos\theta = \dfrac{a}{\sqrt{r^2+a^2}}.

Step 4 — net field: …

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