Why can we treat three charges as a dipole when r >> a?

[gracy]

in this video from time 1:20 to 1:24

he says we can treat the system of three charges as a dipole ignoring middle charge .I want to know why ?I mean just because r>>a how we can ignore the middle charge ?How is it relevant?

 

[jedishrfu]

From your diagram, it appears you may be getting confused about the charge geometry. In the video, the rightmost charge is positive.

 

[gracy]

From your diagram, it appears you may be getting confused about the charge geometry. In the video, the rightmost charge is positive.

No.I know it is positive.
Let me correct it

 

[jedishrfu]

In computing the electric field at P, the video considers the middle charge and says it points upward. The rightmost charge points upward and to the left. The leftmost charge the electric field points downward and to the left so if you draw that out you'll be able to find force on P.

 

[gracy]

But my question is regarding dipole.

 

[blue_leaf77]

During the time interval you mentioned, he temporarily omitted the middle charge because he is calculating the contribution from the other two charge, which turns out to form a dipole. You should see toward the end of the video he stated that the net field would be the total field from the middle charge and the dipole.

 

[gracy]

he stated that the net field would be the total field from the middle charge and the dipole.

yes.But we can consider (the two charges the right most and the leftmost )as a dipole ,right?

 

[blue_leaf77]

But we can consider (the two charges the right most and the leftmost )as a dipole ,right?

Isn't that what the video demonstrates?

 

[gracy]

Isn't that what the video demonstrates?

Wanted to verify

 

[blue_leaf77]

Yes we can.

 

[gracy]

Yes we can.

Any conditions for doing that?I mean is it necessary r>>a ?

 

[blue_leaf77]

No, it applies for any value of $r$. The condition of $r>>a$ was imposed in order for the expression of the field due to the dipole to be able to be written in terms of $r$, instead of the distance from the right (or left) most charge to the observation point.