Q.

(a). Derive the expression for electric field at a point on the equatorial line of an electric dipole. 

(b) Depict the orientation of the dipole in 

(i) stable,           

(ii) unstable equilibrium in a uniform electric field.

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Detailed Solution

Electric dipole moment: It is the product of the magnitude of either charge and distance and distance between them

q=q×2l

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it is a vector quantity whose direction is from negative to positive charge.

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Electric field intensity at P due to +q charge is

E+=14π0qBP2 along PD      =14π0qr2+l2        along PD ...(i)

Electric field intensity at P due to -q charge is

E-=14π0qAP2 along PC      =14π0qr2+l2        along PC ...(ii)

From (i) and (ii), E+=E-=14π0qr2+l2    ...(iii)Net electric field intensity due to the electric dipole at point P

  E=E+2 + E-2 - 2E+E-cos2θ  E=E+2 + E-2 + 2E+2cos2θ              (E+=E-)  E=2E+2 + E+2cos2θ  E=2E+2(1+cos2θ)  E=2E+22cos2θ                    1+cos 2θ=2 cos2θ  E=2E+ cos θ=2×14π0qr2+l2 cos θ                 Using equation   (iii)        Now from OAP, cosθ=1r2+l2       E=2×14π0qr2+l2×lr2+l21/2  E=q×2l4π0r2+l23/2 Since q×2l=p   ...(p is dipole moment)

     E=p4π0r2+l23/2  along (-)x-axis If l << r i.e. dipole is short, then l2 can be negleted as compared to r2 Hence E=p4π0r3   along (-)x-axis

(ii) (a) For stable equilbrium, the angle between p and E is 0,

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    (b) For unstable equilibrium, the angle between p and E is 180,

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For unstable equilibrium, the angle between p and E is 180,

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