Dipole moment of electrically neutral system of charges.

The key is that electric dipole moment is a vector quantity. As such we need to represent each of the charges as a position vector times the corresponding charge at that position. Since the problem tells us that the answer is independent of the origin of the coordinate system, we place the origin in the center of the bottom leg of the equilateral triangle with the -2q charge at the top vertex.

This yields the convenient vectors:

$-2q < 0, \sqrt{3}\frac {d}{2} >$

$q < \frac {d}{2}, 0 >$

$q < \frac {-d}{2}, 0 >$

where q is $1 \mu C = 10^{-6} C$ and d is $1 \mu m = 10^{-6} m$ .

The x-components cancel and only the y-component of the first vector is left giving a vector with magnitude $\sqrt {3}qd = (\sqrt {3})(10^{-6})(10^{-6}) = \sqrt{3} \times 10^{-12}$

or, written as a decimal,

$1.732 \times 10^{-12} C \cdot m$

Since the dipole moment for the system of charges is independent of origin when the system is neutral, we can take one of the vertices of the equilateral triangle as the origin. Now for simplicity assume that the charge $-2Q$ is at origin so that it's position vector has magnitude equal to zero. Then problem reduces to finding resultant of of two vectors with magnitudes $1 \mu m$ each since both the remaining charges have same magnitude. Now it is easy to see that resultant has magnitude equal to $2\times cos 30^\circ \times 10^{-6}$ m. Finally we multiply it with $10^{-6} C$ to get the resultant dipole moment equal to $1.732\times 10^{-12} m.C$ .

the charge on c is seperately divided into -Q+(-Q).one dipole of AC(Q charge of A and -Q charge of C) acts and another along BC(both towards C).As dipole is a vector, p=sqrt(p(square)+p(square)+2p(square)cos 60). here angle is 60(equilateral triangle). p=ql so,sqrt(3)*l=1.732E-12

Dipole moment is the apparent charge distribution from far away from it. Hence we can imagine the two positive charges to collapse int one another as a single particle of magnitude 1 $\mu$ C and positioned between the two positive charges. Now the question demands us to calculate the dipole moment of the apparent positive charge and the actual negative charge. Since the distance between them is $\frac{\sqrt{3}a}{2}$ . Hence Dipole moment can be calculated as to be $\sqrt{3}qa$ .

The magnitude of the vector from one of the 1 μC charges to the -2 μC charge is 1 μm. We can basically just treat the two 1 μC charges as one 2 μC charge which has the same component in the direction of the -2 μC charge (the component perpendicular to the line between the two 1 μC charges) as each of the original 1 μC charges, since the components parallel to the line between the two 1 μC charges cancel out. Therefore, the component is \sqrt{3}/2 μm, and this is the value for r k that we use. We plug this into the formula d=q kr_k, getting 2 μC \times \sqrt{3}/2 μm = \sqrt{3} p m⋅C, or about 1.73 p m⋅C. We must convert to m⋅C, so the answer is 1.73E-12 m⋅C .