What Is the Dipole Moment of This Charge Configuration?

[Saitama]

Homework Statement

4 charges are placed each at a distance 'a' from origin. The dipole moment of the configuration is:

a)2qa$\hat{j}$
b)3qa$\hat{j}$
c)2qa[$\hat{i}+\hat{j}$]
d)none

Homework Equations

The Attempt at a Solution

How do i determine dipole moment here when there is no dipole present?
Dipole consists of two charges equal in magnitude and opposite in sign but i don't see such situation here.

 

[Infinitum]

You can hypothetically assume that the charges are in dipole form. What I mean is, the -2q charge on negative x-axis can be split(hypothetically) as -q and -q charges at the same point. This creates a dipole with the negative y axes +q charge. The 3q charge on y-axis can be split as 2q and q, do you see the dipoles now?

 

[Saitama]

You can hypothetically assume that the charges are in dipole form. What I mean is, the -2q charge on negative x-axis can be split(hypothetically) as -q and -q charges at the same point. This creates a dipole with the negative y axes +q charge. The 3q charge on y-axis can be split as 2q and q, do you see the dipoles now?

Thanks for the reply Infinitum!
I do see the dipoles now, i will get back with a solution.

 

[ehild]

Infinitum is right, you can split the charges so as they form dipoles. But there is a definition of the dipole momentum for a charge distribution. For a system of point charges it is P=∑q_ir_i, where r_i means the position vector of the i-th point charge. You can count the position vectors from the centre of the square. So the position vector of the -2q charge on the right is ai, that of the 3q charge is aj, and so on. The result is the same that you get with the splitting method.


ehild

 

[Saitama]

Thank you both Infinitum and ehild!
I get my answer to be 2qa$\hat{j}$. ehild, thank you for the formula, that might come in handy because i have similar problems like this one.

 

[ehild]

You will learn that formula sooner or later.

ehild