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Physics · Electrostatics

JEE Advanced 2018 — Paper 2 — Question 14

The electric field E is measured at a point P(0,0, d)\mathrm{P}(0,0, \mathrm{~d}) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between EE and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II.

List-IList-II
P. EE is independent of d1. A point charge Q at the origin
Q. EE ∝1d\propto \frac{1}{\text{d}}2. A small dipole with point charges Q at (0,0,ℓ)(0,0, \ell) and -Q at (0,0,−ℓ)(0,0,-\ell). Take 2ℓ≪ d2 \ell \ll \mathrm{~d}
R. E∝1 d2 \mathrm{E} \propto \frac{1}{\mathrm{~d}^{2}}3. An infinite line charge coincident with the xx-axis, with uniform linear charge density λ\lambda
S. E∝1 d3 \mathrm{E} \propto \frac{1}{\mathrm{~d}^{3}}4. Two infinite wires carrying uniform linear charge density parallel to the xx - axis. The one along (y=0,z=ℓ)(y=0, z=\ell) has a charge density +λ+\lambda and the one along (y=0,z=−ℓ)(\mathrm{y}=0, \mathrm{z}=-\ell) has a charge density −λ-\lambda. Take 2ℓ≪d2 \ell \ll d
5. Infinite plane charge coincident with the xyx y-plane with uniform surface charge density
  1. Option A:

    P→5;Q→3,4;R→1;S→2\mathrm{P} \rightarrow 5 ; \quad \mathrm{Q} \rightarrow \mathbf{3}, \mathbf{4} ; \quad \mathrm{R} \rightarrow \mathbf{1} ; \quad \mathrm{S} \rightarrow \mathbf{2}

  2. Option B:

    P→5;Q→3;R→1,4;S→2\mathbf{P} \rightarrow 5 ; \quad \mathrm{Q} \rightarrow 3 ; \quad \mathrm{R} \rightarrow 1,4 ; \quad \mathrm{S} \rightarrow 2

    Correct
  3. Option C:

    P→5;Q→3;R→1,2;S→4\mathrm{P} \rightarrow 5 ; \quad \mathrm{Q} \rightarrow 3 ; \quad \mathrm{R} \rightarrow 1,2 ; \quad \mathrm{S} \rightarrow 4

  4. Option D:

    P→4;Q→2,3;R→1;S→5\mathbf{P} \rightarrow \mathbf{4} ; \quad \mathrm{Q} \rightarrow \mathbf{2}, \mathbf{3} ; \quad \mathrm{R} \rightarrow \mathbf{1} ; \quad \mathrm{S} \rightarrow \mathbf{5}

Answer: B

Step-by-step solution

For point charge, E∝1r2E \propto \frac{1}{r^{2}}

For infinite line charge, E∝1r\mathrm{E} \propto \frac{1}{\mathrm{r}}

For infinite plane E=\mathrm{E}= constant

For small dipole E∝1r3E \propto \frac{1}{r^{3}}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Field & motion of charge

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