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Electrostatics question

2020 · 4 Sep · Shift 1 · Q49
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Electrostatics question

2020 · 4 Sep · Shift 1 · Q49

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A two point charges 4q and -q are fixed on the x-axis at x = −d2- {d \over 2}−2d​ and x =d2{d \over 2}2d​ respectively. If a third point charge 'q' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will : JEE Main 2020 (Online) 4th September Morning Slot Physics - Electrostatics Question 174 English
  1. A
    increase by 3q24πε0d{{3{q^2}} \over {4\pi {\varepsilon _0}d}}4πε0​d3q2​
  2. B
    increase by 2q23πε0d{{2{q^2}} \over {3\pi {\varepsilon _0}d}}3πε0​d2q2​
  3. C
    decrease by q24πε0d{{{q^2}} \over {4\pi {\varepsilon _0}d}}4πε0​dq2​
  4. D
    decrease by 4q23πε0d{{4{q^2}} \over {3\pi {\varepsilon _0}d}}3πε0​d4q2​
View written solutionFree

Correct answer: D

  1. Use electrostatic potential energy change

The change in potential energy of the moving charge qqq depends only on the initial and final positions:

ΔU=q (Vf−Vi)\Delta U = q\,(V_f - V_i)ΔU=q(Vf​−Vi​)

where ViV_iVi​ and VfV_fVf​ are the potentials due to the fixed charges 4q4q4q and −q-q−q.

  1. Initial position: origin x=0x=0x=0

Fixed charges are at:

  • 4q4q4q at x=−d2x=-\dfrac d2x=−2d​
  • −q-q−q at x=d2x=\dfrac d2x=2d​

Distance of origin from each charge is d2\dfrac d22d​.

So,

Vi=14πε0(4qd/2+−qd/2)V_i = \frac{1}{4\pi\varepsilon_0}\left(\frac{4q}{d/2} + \frac{-q}{d/2}\right)Vi​=4πε0​1​(d/24q​+d/2−q​) Vi=14πε0(8qd−2qd)=14πε0⋅6qdV_i = \frac{1}{4\pi\varepsilon_0}\left(\frac{8q}{d} - \frac{2q}{d}\right) = \frac{1}{4\pi\varepsilon_0}\cdot \frac{6q}{d}Vi​=4πε0​1​(d8q​−d2q​)=4πε0​1​⋅d6q​

Hence initial energy of the moving charge qqq is

Ui=qVi=14πε0⋅6q2dU_i = qV_i = \frac{1}{4\pi\varepsilon_0}\cdot \frac{6q^2}{d}Ui​=qVi​=4πε0​1​⋅d6q2​
  1. Final position: x=dx=dx=d

Now find distances from x=dx=dx=d to the fixed charges.

  • Distance from 4q4q4q at x=−d/2x=-d/2x=−d/2:
r_1 = d - \left(-\frac d2\right) = \frac{3d}{2}$$ - Distance from $-q$ at $x=d/2$:

r_2 = d - \frac d2 = \frac d2$$

Therefore,

Vf=14πε0(4q3d/2+−qd/2)V_f = \frac{1}{4\pi\varepsilon_0}\left(\frac{4q}{3d/2} + \frac{-q}{d/2}\right)Vf​=4πε0​1​(3d/24q​+d/2−q​) Vf=14πε0(8q3d−2qd)V_f = \frac{1}{4\pi\varepsilon_0}\left(\frac{8q}{3d} - \frac{2q}{d}\right)Vf​=4πε0​1​(3d8q​−d2q​) Vf=14πε0(8q−6q3d)=14πε0⋅2q3dV_f = \frac{1}{4\pi\varepsilon_0}\left(\frac{8q-6q}{3d}\right) = \frac{1}{4\pi\varepsilon_0}\cdot \frac{2q}{3d}Vf​=4πε0​1​(3d8q−6q​)=4πε0​1​⋅3d2q​

Hence final energy is

Uf=qVf=14πε0⋅2q23dU_f = qV_f = \frac{1}{4\pi\varepsilon_0}\cdot \frac{2q^2}{3d}Uf​=qVf​=4πε0​1​⋅3d2q2​
  1. Change in energy
ΔU=Uf−Ui\Delta U = U_f - U_iΔU=Uf​−Ui​ ΔU=14πε0(2q23d−6q2d)\Delta U = \frac{1}{4\pi\varepsilon_0}\left(\frac{2q^2}{3d} - \frac{6q^2}{d}\right)ΔU=4πε0​1​(3d2q2​−d6q2​)

Take common denominator 3d3d3d:

ΔU=14πε0(2q2−18q23d)=14πε0(−16q23d)\Delta U = \frac{1}{4\pi\varepsilon_0}\left(\frac{2q^2 - 18q^2}{3d}\right) = \frac{1}{4\pi\varepsilon_0}\left(\frac{-16q^2}{3d}\right)ΔU=4πε0​1​(3d2q2−18q2​)=4πε0​1​(3d−16q2​) ΔU=−16q212πε0d=−4q23πε0d\Delta U = -\frac{16q^2}{12\pi\varepsilon_0 d} = -\frac{4q^2}{3\pi\varepsilon_0 d}ΔU=−12πε0​d16q2​=−3πε0​d4q2​

So the energy decreases by

4q23πε0d\frac{4q^2}{3\pi\varepsilon_0 d}3πε0​d4q2​
  1. Option check
  • A: incorrect
  • B: incorrect
  • C: incorrect
  • D: correct

Thus, the correct answer is D.

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