# Two charges q,-q are placed at a distance apart . what is the point where the resultant field is parallel to the line joining them.

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2015-05-17T01:38:29+05:30

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See diagram.

When an electrical dipole of charges  q  at A and - q at B are  kept at a distance  2 * a  apart, there is an electric field in the space around them.

There are various points on the axis connecting the dipole: R, S, T , and U.  At all these points, the electric field due to the two charges q and -q are as follows.  The distance between the point in consideration and the charges be d1 and d2 respectively.

E  =  E1  +  E2
=  1/(4πε) * q/d₁²  - 1/(4πε)  q/d₂²
E = q / (4πε)  [1/d₁²  - 1/d₂² ]

For all the points R, S T and U along the dipole axis AB, the E1 and E2 are vectors parallel to AB.  Hence, the resultant is also parallel to the axis AB.

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See the points P and Q on the perpendicular bisector of the axis AB.

The field due to q at A is in the direction E1 as shown.  The electric field due to -q at B is in the direction E2 as shown.  Let AP = BP = d.  These two fields are equal in magnitude as:

magnitudes   E1 = 1/(4πε) * q/d²  = E2

Component of electric field E1 along the horizontal = E1 Cos Ф = E1 * a / d
component of electric field E2 along the horizontal = E2 Cos Ф = E2 * a / d

The components of E1 and E2 along the vertical directions (E1 Sin Ф and E2 sinФ) cancel each other.

Hence, the resultant electric field vector at P = 2 * 1/(4πε) * q /d² * a / d

E =  1/(2πε) * q a / d³

At all points on the perpendicular bisector of the dipole, the resultant electric field is parallel to the axis.

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