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Electrostatics question

2018 · Shift 2 · Q51
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Electrostatics question

2018 · Shift 2 · Q51

JEE AdvancedPhysicsElectrostaticsMCQ+3 / −0.75
The electric field EEE is measured at a point P(0,0,d)P(0,0,d)P(0,0,d) generated due to various charge distributions and the dependence of EEE on ddd is found to be different for different charge distributions. List-I{\rm I}I contains different relations between EEE and ddd. List-II{\rm II}II describes different electric charge distributions, along with their locations. Match the functions in List-I{\rm I}I with the related charge distributions in List-II{\rm II}II.

LIST - I LIST - II
P. EEE is independent of ddd 1. A point charge Q at the origin
Q. E ∝ 1/dE\, \propto \,1/dE∝1/d 2. A small dipole with point charges
QQQ at (0,0,l)\left( {0,0,l} \right)(0,0,l) and −Q-Q−Q at
(0,0,−l).\left( {0,0, - l} \right).(0,0,−l). Take 2l<<d2l \lt \lt d2l<<d
R. E ∝ 1/d2E\, \propto \,1/{d^2}E∝1/d2 3. An infinite line charge coincident
with the x-axis, with uniform linear charge density λ\lambdaλ
S. E ∝ 1/d3E\, \propto \,1/{d^3}E∝1/d3 4. Two infinite wires carrying
uniform linear charge density
parallel to the xxx-axis. The one
along (y=0,z=l)\left( {y = 0,z = l} \right)(y=0,z=l) has
a charge density +λ+ \lambda+λ and the one
along (y=0,z=−l)\left( {y = 0,z = - l} \right)(y=0,z=−l) has a
charge density Take
5. Infinite plane charge coincident
with the xyxyxy-plane with uniform surface charge density
  1. A
    P→5;Q→3,4;R→1;S→2P \to 5;Q \to 3,4;R \to 1;S \to 2P→5;Q→3,4;R→1;S→2
  2. B
    P→5;Q→3;R→1,4;S→2P \to 5;Q \to 3;R \to 1,4;S \to 2P→5;Q→3;R→1,4;S→2
  3. C
    P→4;Q→3;R→1,2;S→4P \to 4;Q \to 3;R \to 1,2;S \to 4P→4;Q→3;R→1,2;S→4
  4. D
    P→4;Q→2,3;R→1;S→5P \to 4;Q \to 2,3;R \to 1;S \to 5P→4;Q→2,3;R→1;S→5
View written solutionFree

Correct answer: B

  1. Interpret the charge distributions

We need the electric field at P(0,0,d)P(0,0,d)P(0,0,d) and its dependence on ddd.

List-II contains:

  1. Point charge QQQ at origin.
  2. Small dipole with charges +Q+Q+Q at (0,0,l)(0,0,l)(0,0,l) and −Q-Q−Q at (0,0,−l)(0,0,-l)(0,0,−l), with 2l≪d2l \ll d2l≪d.
  3. Infinite line charge along the xxx-axis with linear charge density λ\lambdaλ.
  4. Two infinite line charges parallel to the xxx-axis, one at z=lz=lz=l with +λ+\lambda+λ and one at z=−lz=-lz=−l with −λ-\lambda−λ.
  5. Infinite plane sheet in the xyxyxy-plane with surface charge density σ\sigmaσ.

We now find how EEE varies with ddd in each case.


  1. Case 1: Point charge at origin

Distance of P(0,0,d)P(0,0,d)P(0,0,d) from origin is ddd.

E=14πε0Qd2E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{d^2}E=4πε0​1​d2Q​

So,

E∝1d2E\propto \frac{1}{d^2}E∝d21​

Hence,

1→R1 \to R1→R


  1. Case 2: Small dipole on the zzz-axis

Point PPP lies on the axial line of the dipole. For a short dipole, axial field is

E=14πε02pd3,p=Q(2l)E = \frac{1}{4\pi\varepsilon_0}\frac{2p}{d^3}, \qquad p=Q(2l)E=4πε0​1​d32p​,p=Q(2l)

Thus,

E∝1d3E\propto \frac{1}{d^3}E∝d31​

Hence,

2→S2 \to S2→S


  1. Case 3: Infinite line charge along the xxx-axis

The perpendicular distance of P(0,0,d)P(0,0,d)P(0,0,d) from the xxx-axis is ddd.

Field due to an infinite line charge is

E=λ2πε0dE=\frac{\lambda}{2\pi\varepsilon_0 d}E=2πε0​dλ​

So,

E∝1dE\propto \frac{1}{d}E∝d1​

Hence,

3→Q3 \to Q3→Q


  1. Case 4: Two infinite line charges at z=±lz=\pm lz=±l

One line at z=lz=lz=l has +λ+\lambda+λ, the other at z=−lz=-lz=−l has −λ-\lambda−λ.

At P(0,0,d)P(0,0,d)P(0,0,d), distances from the two lines are:

r+=d−l,r−=d+lr_+ = d-l, \qquad r_- = d+lr+​=d−l,r−​=d+l

Field magnitudes:

E+=λ2πε0(d−l),E−=λ2πε0(d+l)E_+ = \frac{\lambda}{2\pi\varepsilon_0(d-l)}, \qquad E_- = \frac{\lambda}{2\pi\varepsilon_0(d+l)}E+​=2πε0​(d−l)λ​,E−​=2πε0​(d+l)λ​

Their directions are opposite along the zzz-axis, so net field magnitude is

E=λ2πε0(1d−l−1d+l)E = \frac{\lambda}{2\pi\varepsilon_0}\left(\frac{1}{d-l}-\frac{1}{d+l}\right)E=2πε0​λ​(d−l1​−d+l1​)

E=λ2πε0⋅2ld2−l2E = \frac{\lambda}{2\pi\varepsilon_0}\cdot \frac{2l}{d^2-l^2}E=2πε0​λ​⋅d2−l22l​

For d≫ld \gg ld≫l,

E≈λ2πε0⋅2ld2∝1d2E \approx \frac{\lambda}{2\pi\varepsilon_0}\cdot \frac{2l}{d^2} \propto \frac{1}{d^2}E≈2πε0​λ​⋅d22l​∝d21​

So this behaves like

E∝1d2E\propto \frac{1}{d^2}E∝d21​

Hence,

4→R4 \to R4→R


  1. Case 5: Infinite plane sheet in the xyxyxy-plane

Field due to an infinite plane sheet is independent of distance:

E=σ2ε0E=\frac{\sigma}{2\varepsilon_0}E=2ε0​σ​

So,

E is independent of dE \text{ is independent of } dE is independent of d

Hence,

5→P5 \to P5→P


  1. Now match List-I with List-II
  • PPP: EEE independent of ddd →5\to 5→5
  • QQQ: E∝1/dE\propto 1/dE∝1/d →3\to 3→3
  • RRR: E∝1/d2E\propto 1/d^2E∝1/d2 →1,4\to 1,4→1,4
  • SSS: E∝1/d3E\propto 1/d^3E∝1/d3 →2\to 2→2

So the correct matching is:

P→5,Q→3,R→1,4,S→2P\to 5,\quad Q\to 3,\quad R\to 1,4,\quad S\to 2P→5,Q→3,R→1,4,S→2

This corresponds to Option B.


  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They match.

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