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# Two equal charges ‘q′ of opposite sign are separated by a small distance ‘d′. The electric intensity ‘E′ at a point on the straight line passing through the two charges at a very large distance ‘r′ from the midpoint of two charges is :14πε0qdr214πε02qdr214πε0qdr314πε02qdr3

A
14πε0qdr2
B
14πε02qdr2
C
14πε02qdr3
D
14πε0qdr3
Solution
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#### As shown in the fig there are two electric fields E+q and E−q due to +q and −q charges respectively.Resultant of this two oppositely directed electric fields gives the net electric field at that point.∴ Electric field at that point is given by,E=E+q−E−q=Kq/(r−d/2)2−Kq/(r+d/2)2 =Kq[1(r2−rd+d4/4)−1(r2+rd+d4/4)] =Kq[1(r2−rd)−1(r2+rd)] ......(d2/4≈0) =Kq[(r2+rd)r4−r2d2−(r2−rd)r4−r2d2] =Kq[(2rd)r2(r2−d2)] =Kq(2rd)r4 ....(d2≈0) ∴E=14πϵo2qdr3

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