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Physics Question on Electric Field

Two point charges of 1 μC1 \,\mu C and −1 μC-1\, \mu C are separated by a distance of 100 ?100\, ?. A point PP is at a distance of 10 cm10 \,cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at PP will be

A

9 N C−19 \,N \,C^{-1}

B

0.9 N C−10.9 \,N \,C^{-1}

C

90 N C−190 \,N \,C^{-1}

D

0.09 N C−10.09 \,N \,C^{-1}

Answer

0.09 N C−10.09 \,N \,C^{-1}

Explanation

Solution

The point lies on equatorial line of a short dipole. ∴E=2ql4πε0r3\therefore E=\frac{2ql}{4\pi\varepsilon_{0}r^{3}} =9×109×10−6×10−8(10−1)3=\frac{9\times10^{9}\times10^{-6}\times10^{-8}}{\left(10^{-1}\right)^{3}} =0.09 N C−1=0.09\,N\,C^{-1}