# 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.

1
by aditipandey

Log in to add a comment

by aditipandey

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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.*

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

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