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Electrostatics question

2021 · 22 Jul · Shift 2 · Q54
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  5. /2021 · 22 Jul · Shift 2 · Q54

Electrostatics question

2021 · 22 Jul · Shift 2 · Q54

JEE MainPhysicsElectrostaticsMCQ+4 / −1
An electric dipole is placed on x-axis in proximity to a line charge of linear charge density 3.0 ×\times× 10 −-− 6 C/m. Line charge is placed on z-axis and positive and negative charge of dipole is at a distance of 10 mm and 12 mm from the origin respectively. If total force of 4N is exerted on the dipole, find out the amount of positive or negative charge of the dipole.
  1. A
    0.485 mC
  2. B
    815.1 nC
  3. C
    8.8 μ\muμ C
  4. D
    4.44 μ\muμ C
View written solutionFree

Correct answer: D

  1. Electric field due to an infinite line charge

A line charge of linear charge density λ=3.0×10−6 C/m\lambda = 3.0\times 10^{-6}\,\text{C/m}λ=3.0×10−6C/m lies along the zzz-axis.

At a perpendicular distance rrr from an infinite line charge, the electric field is

E=λ2πε0r=2kλrE = \frac{\lambda}{2\pi \varepsilon_0 r} = \frac{2k\lambda}{r}E=2πε0​rλ​=r2kλ​

where k=9×109 N m2/C2k = 9\times 10^9\,\text{N m}^2/\text{C}^2k=9×109N m2/C2.

So,

E=2(9×109)(3×10−6)r=54×103rE = \frac{2(9\times 10^9)(3\times 10^{-6})}{r} = \frac{54\times 10^3}{r}E=r2(9×109)(3×10−6)​=r54×103​

with rrr in meters.

  1. Positions of the dipole charges

The dipole is on the xxx-axis.

  • Positive charge is at distance 10 mm=0.01 m10\,\text{mm} = 0.01\,\text{m}10mm=0.01m from origin.
  • Negative charge is at distance 12 mm=0.012 m12\,\text{mm} = 0.012\,\text{m}12mm=0.012m from origin.

Since the line charge is on the zzz-axis, these distances are the perpendicular distances from the line charge.

Thus,

E+=54×1030.01=5.4×106 N/CE_+ = \frac{54\times 10^3}{0.01} = 5.4\times 10^6\,\text{N/C}E+​=0.0154×103​=5.4×106N/C E−=54×1030.012=4.5×106 N/CE_- = \frac{54\times 10^3}{0.012} = 4.5\times 10^6\,\text{N/C}E−​=0.01254×103​=4.5×106N/C
  1. Forces on the dipole charges

Let magnitude of each dipole charge be qqq.

Force on positive charge:

F+=qE+F_+ = qE_+F+​=qE+​

Force on negative charge:

F−=qE−F_- = qE_-F−​=qE−​

These two forces act in opposite directions because the charges are opposite in sign.

Hence net force on the dipole is

Fnet=q(E+−E−)F_{\text{net}} = q(E_+ - E_-)Fnet​=q(E+​−E−​)

Given Fnet=4 NF_{\text{net}} = 4\,\text{N}Fnet​=4N,

4=q(5.4×106−4.5×106)4 = q(5.4\times 10^6 - 4.5\times 10^6)4=q(5.4×106−4.5×106) 4=q(0.9×106)4 = q(0.9\times 10^6)4=q(0.9×106) q=40.9×106q = \frac{4}{0.9\times 10^6}q=0.9×1064​ q=4.44×10−6 Cq = 4.44\times 10^{-6}\,\text{C}q=4.44×10−6C q=4.44 μCq = 4.44\,\mu\text{C}q=4.44μC
  1. Option check

The value of charge is

4.44 μC\boxed{4.44\,\mu\text{C}}4.44μC​

This matches Option D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.

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