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Two charges +20muC and -20muC are placed...

Two charges `+20muC and -20muC` are placed 10mm apart. The electric field at point P, on the axis of the dipole 10 cm away from its centre O on the side of the positive charge is

A

`8.6xx10^(9)NC^(-1)`

B

`4.1xx10^(6)NC^(-1)`

C

`3.6xx10^(6)NC^(-1)`

D

`4.6xx10^(5)NC^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C

Here, `q=pm20muC=pm20xx10^(-6)C,2a=10mm=10xx10^(-3)m`
`r=OP=10cm=10xx10^(-2)m`
`|vecp|=qxx2a=20xx10^(-6)xx10xx10^(-3)m=2xx10^(-7)m`
The electirc field along BP, `vecE=(2vecpr)/(4piepsilon_(0)(r^(2)-a^(2))^(2))`
As" " a lt lt r,
`vecE=(2|vecp|)/(4piepsilon_(0)r^(3))=(2xx2xx10^(-7)xx9xx10^(9))/((10xx10^(-2))^(3))=3.6xx10^(6)NC^(_1)`
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