Question

In the following configuration of charges. Find the net dipole moment of the system :

• A. \(\sqrt{180}\) qa
• B. \(\sqrt{150}\) qa
• C. \(\sqrt{200}\) qa
• D. \(\sqrt{140}\) qa

Hint:

For a system of charges, first check if the net charge is zero. If it is, you can decompose the system into pairs of dipoles. Splitting a larger charge into smaller ones (e.g., -4q into -2q and -2q) is a very effective strategy to create simple dipole pairs. The final net dipole moment is the vector sum of the moments of these pairs.

Answer 1

The Correct Option is A

Step 1: Physical idea used

The electric dipole moment of a system of point charges is obtained by adding the products of each charge and its position vector.

Since the total charge of the system is zero, the dipole moment does not depend on the choice of origin.

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Step 2: Choosing the reference origin

We choose the origin at point O with coordinates (0, 0).

Positions of charges:

Charge minus 4q is at (-2a, 0)
Charge plus 2q is at (2a, 0)
Charge plus 2q is at (0, -3a)

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Step 3: Dipole moment contribution of each charge

Dipole moment contribution equals charge multiplied by its position.

For charge minus 4q:

Contribution = minus 4q multiplied by (-2a, 0)

Contribution = (8qa, 0)

For charge plus 2q at (2a, 0):

Contribution = (4qa, 0)

For charge plus 2q at (0, -3a):

Contribution = (0, -6qa)

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Step 4: Net dipole moment

Net dipole moment is obtained by adding all contributions.

X component = 8qa plus 4qa = 12qa
Y component = 0 minus 6qa = minus 6qa

So the dipole moment vector is:

(12qa, -6qa)

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Step 5: Magnitude of dipole moment

Magnitude is obtained using Pythagoras theorem.

Magnitude equals square root of (12qa squared plus 6qa squared)

Magnitude equals square root of 180 multiplied by q multiplied by a

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Final Answer:

The magnitude of the net electric dipole moment is
square root of 180 multiplied by q multiplied by a