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Electrostatics question

2019 · 12 Jan · Shift 1 · Q61
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Electrostatics question

2019 · 12 Jan · Shift 1 · Q61

JEE MainPhysicsElectrostaticsMCQ+4 / −1
Determine the electric dipole moment of the system of the three charges, placed on the vertices of an equilateral triangle, as shown in the figure : JEE Main 2019 (Online) 12th January Morning Slot Physics - Electrostatics Question 196 English
  1. A
    2qℓj^2q\ell \widehat j2qℓj​
  2. B
    (qℓ)i^+j^2\left( {q\ell } \right){{\widehat i + \widehat j} \over {\sqrt 2 }}(qℓ)2​i+j​​
  3. C
    3 qℓj^−i^2\sqrt 3 \,q\ell {{\widehat j - \widehat i} \over {\sqrt 2 }}3​qℓ2​j​−i​
  4. D
    −3 qℓj^- \sqrt 3 \,q\ell \widehat j−3​qℓj​
View written solutionFree

Correct answer: D

  1. Use the definition of dipole moment

For a system of point charges,

p⃗=∑iqir⃗i\vec p = \sum_i q_i \vec r_ip​=i∑​qi​ri​

where r⃗i\vec r_iri​ is the position vector of each charge with respect to the chosen origin.


  1. Take the geometry of the equilateral triangle

Let the side of the equilateral triangle be ℓ\ellℓ.

For an equilateral triangle, the height is

h=32ℓh = \frac{\sqrt{3}}{2}\ellh=23​​ℓ

From the standard figure for this question, the two equal positive charges at the base and one negative charge at the top give a net charge zero, so the dipole moment is origin-independent.

Choose origin at the midpoint of the base, with:

  • base along the xxx-axis,
  • upward direction along j^\hat jj^​.

Then coordinates are:

  • left base vertex: (−ℓ2,0)\left(-\frac{\ell}{2},0\right)(−2ℓ​,0) with charge +q+q+q
  • right base vertex: (ℓ2,0)\left(\frac{\ell}{2},0\right)(2ℓ​,0) with charge +q+q+q
  • top vertex: (0,32ℓ)\left(0,\frac{\sqrt3}{2}\ell\right)(0,23​​ℓ) with charge −2q-2q−2q

  1. Compute contribution of each charge

For the two base charges:

q(−ℓ2i^)+q(ℓ2i^)=0q\left(-\frac{\ell}{2}\hat i\right) + q\left(\frac{\ell}{2}\hat i\right)=0q(−2ℓ​i^)+q(2ℓ​i^)=0

Their xxx-components cancel, and they have no yyy-component.

For the top charge:

(−2q)(32ℓj^)=−3 qℓ j^(-2q)\left(\frac{\sqrt3}{2}\ell\hat j\right) = -\sqrt3\,q\ell\,\hat j(−2q)(23​​ℓj^​)=−3​qℓj^​

Thus,

p⃗=−3 qℓ j^\vec p = -\sqrt3\,q\ell\,\hat jp​=−3​qℓj^​
  1. Match with the options

This corresponds to

−3 qℓ j^-\sqrt3\,q\ell\,\hat j−3​qℓj^​

which is Option D.


  1. Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So they agree.

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