Electric Dipole- An electric dipole is a pair of equal and opposite point charges q and –q separated by a distance ‘2a’ . By convention , the direction from –q to q is said to be direction of the dipole moment (p).
Case 1 - For point on the axis
Let the point P be at distance r from the centre of the dipole of charge
E-q = pq/4Пє0(r+a)2 ---------------------- (1)
Where p is the vector along the dipole axis (from –q to q) also
E+q = pq/4Пє0(r-a)2 -----------------------(2)
The total field at P is
E = E+q + E-q
E= pq/4Пє0[1/(r-a)2 – 1/(r+a)2]
= pq/4Пє0 × 4ar/(r2 – a2)2 (For r > > > a )
E = 4qa/ 4Пє0r3
Case 2- For points on the equatorial plane
The magnitudes of the electric field due to the two charges +q and –q are given by
E+q= q/4Пє0 ×1/(r2 + a2)
E-q= q/4Пє0 ×1/ (r2 + a2)
The total electric field is opposite to p^
E = - (E+q + E-q)cosө p^
= 2qa/4Пє0 (r2 + a2)3/2 p^
At large distances (r > > > a) , this reduces to
E =2qa/4Пє0 r3 p^ (r > > > a)
P = q× 2a p^ Where p = electric dipole moment
E = 2p/4Пє0 r3 (at a point on the axis ) (r > > > a)
At a point on the equatorial plane
E = p/4Пє0 r3 (r > > a)
dipole in uniform external field
magnitude of torque = force×┴ distance
=qE×2a.sinө
=(q×2a)E.sinө
=pE.sinө
T=p×E
FIELD DUE TO UNIFORMLY CHARGED INFINITE PLANE SHEET-
The net flux through the Gaussian surface is 2EA
The charge enclosed by the closed surface is σ A
By applying Gauss’s law
2EA = σA/ є0
Or E = σ /2 є0
In vector form
E = σ /2 є0 n^
E = σ R2/ є0 r2 = q/4Пє0 r2
is it ture for any pastion of point
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