The dipole moment for a system of charges is defined as

d⃗ =∑kqkr⃗ k.
(see here for a more in depth discussion of the dipole moment). Determine the magnitude of the dipole moment in m⋅C for a system of three charges (Q=1 μC, Q=1 μC, and −2Q=−2 μC) located on the vertices of an equilateral triangle of side a=1 μm. One can show that for an electrically neutral system, the dipole moment is independent of the origin of coordinates (which makes this problem unambiguous).

1.732

Shame on you Keshav!!! Cheating on Brilliant!!! This site is meant to be a platform to practice your own skills, not to copy paste the questions and get free answers and then get incentives without effort. So either play fair and be honest or leave this site. People like you are shame to the Brilliant community. And to the others, please give the answer to this problem after Monday 10/6/2013, so that this cheat doesn't get the opportunity to cheat.

I think you don't even understand the problem fully. For you it's a child's game isn't it? Just copy-paste the questions and post them here? You didn't even see if the formatting was correct or not.

yes i support with shame its really a shame .

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wrong

I am much better than u all

To determine the magnitude of the dipole moment for a system of three charges located on the vertices of an equilateral triangle, we can follow these steps:

1. First, we need to calculate the position vector for each charge with respect to the origin of coordinates. Since the charges are located on the vertices of an equilateral triangle, the positions will form a regular triangle.

2. Let's assume that one vertex of the triangle is at the origin (0,0). The other two vertices will be at (a,0) and (a/2, a√3/2), where 'a' is the side length of the triangle.

3. Now, we can calculate the position vectors (r) for each charge. Let's assign q1 = Q, q2 = Q, and q3 = -2Q to the three charges.

- The position vector r1 for charge q1 will be (0,0) since it is at the origin.
- The position vector r2 for charge q2 will be (a,0), and
- The position vector r3 for charge q3 will be (a/2, a√3/2).

4. Next, we need to calculate the magnitude of the dipole moment using the formula:

d = ∑(q*r)

where ∑ denotes the sum over all charges.

Let's substitute the values into the formula:

d = q1*r1 + q2*r2 + q3*r3
= Q*(0,0) + Q*(a,0) + (-2Q)*(a/2, a√3/2)

Simplifying this expression:

d = Q*(a, 0) - Q*(a/2, a√3/2)

Breaking down the vector components:

d = (Q*a - (Q/2)*a, 0 - (Q/2)*a√3)

d = (Q*a/2, -Q*a√3/2)

5. Finally, to determine the magnitude of the dipole moment, we take the magnitude of the resulting vector:

|d| = √[(Q*a/2)^2 + (-Q*a√3/2)^2]

|d| = √[(Q^2*a^2/4) + (Q^2*a^2*3/4)]

|d| = √[(Q^2*a^2 + 3*Q^2*a^2)/4]

|d| = √[4Q^2*a^2/4]

Simplifying,

|d| = √[Q^2*a^2]

|d| = Q*a

Therefore, the magnitude of the dipole moment, in m⋅C, for the given system of three charges is Q*a, which is (1 μC) * (1 μm) = 1 μC⋅μm.