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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 :

- 14πε0qdr2
- 14πε02qdr2
- 14πε0qdr3
- 14πε02qdr3

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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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