# 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.

by aditipandey 14.05.2015

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by aditipandey 14.05.2015

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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.*

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 ]

============

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