JEE Challenger
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PhyMedAdv 2018 • Paper 2

Matching Electric Field Dependence on Distance for Various Charge Distributions

Consider the electric field EE evaluated at point P(0,0,d)P(0,0,d) produced by different charge configurations, where EE depends on distance dd in various ways. List-I specifies functional relations between EE and dd, whereas List-II details various charge distributions and their positions. Pair each entry in List-I with its corresponding charge arrangement from List-II.

LIST-ILIST-IIP. E is independent of d1. A point charge Q located at the originQ. E∝1d2. A small dipole formed by point charges Q at (0,0,l) and −Q at (0,0,−l), with 2l≪dR. E∝1d23. An infinite line charge aligned with the x-axis, having uniform linear charge density λS. E∝1d34. Two infinite parallel wires with uniform linear charge density along the x-axis. The wire at (y=0,z=l)has density +λ and the one at (y=0,z=−l) has density −λ, with 2l≪d5. An infinite planar sheet of charge in the xy-plane with uniform surface charge density\begin{array}{ll} \text{\textbf{LIST-I}} & \text{\textbf{LIST-II}} \\ \text{P. } E \text{ is independent of } d & \text{1. A point charge } Q \text{ located at the origin} \\ \text{Q. } E \propto \frac{1}{d} & \text{2. A small dipole formed by point charges } Q \text{ at } (0,0,l) \text{ and } -Q \text{ at } (0,0,-l), \text{ with } 2l \ll d \\ \text{R. } E \propto \frac{1}{d^2} & \text{3. An infinite line charge aligned with the } x\text{-axis, having uniform linear charge density } \lambda \\ \text{S. } E \propto \frac{1}{d^3} & \text{4. Two infinite parallel wires with uniform linear charge density along the } x\text{-axis. The wire at } (y=0, z=l) \\ & \quad \text{has density } +\lambda \text{ and the one at } (y=0, z=-l) \text{ has density } -\lambda, \text{ with } 2l \ll d \\ & \text{5. An infinite planar sheet of charge in the } xy\text{-plane with uniform surface charge density} \end{array}

Options

A

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

B

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

Correct
C

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

D

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

Topics & Concepts

Electric Charges and FieldsElectric FieldGauss's LawCoulomb's Law

Step-by-Step Solution

To determine the dependence of the electric field EE on the distance dd at point P(0,0,d)P(0,0,d) for each charge distribution in List-II, we analyze them individually:

Analysis of List-II Charge Configurations:

  1. A point charge QQ at the origin (0,0,0)(0,0,0): The distance from the origin to P(0,0,d)P(0,0,d) is dd. By Coulomb's Law, the electric field is: E=14πε0Qd2  ⟹  E∝1d2E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{d^2} \implies E \propto \frac{1}{d^2} This corresponds to entry R. Thus, R→1\text{R} \rightarrow \mathbf{1}.

  2. A small dipole formed by point charges QQ at (0,0,l)(0,0,l) and −Q-Q at (0,0,−l)(0,0,-l) with 2l≪d2l \ll d: The point P(0,0,d)P(0,0,d) lies on the axial line (z-axis) of the dipole. The net electric field is: E=Q4πε0[1(d−l)2−1(d+l)2]=Q4πε0d2[(1−ld)−2−(1+ld)−2]E = \frac{Q}{4\pi\varepsilon_0} \left[ \frac{1}{(d-l)^2} - \frac{1}{(d+l)^2} \right] = \frac{Q}{4\pi\varepsilon_0 d^2} \left[ \left(1-\frac{l}{d}\right)^{-2} - \left(1+\frac{l}{d}\right)^{-2} \right] Using binomial approximation for l≪dl \ll d: E≈Q4πε0d2[(1+2ld)−(1−2ld)]=2Q(2l)4πε0d3  ⟹  E∝1d3E \approx \frac{Q}{4\pi\varepsilon_0 d^2} \left[ \left(1 + \frac{2l}{d}\right) - \left(1 - \frac{2l}{d}\right) \right] = \frac{2Q(2l)}{4\pi\varepsilon_0 d^3} \implies E \propto \frac{1}{d^3} This corresponds to entry S. Thus, S→2\text{S} \rightarrow \mathbf{2}.

  3. An infinite line charge along the xx-axis with uniform linear charge density λ\lambda: The perpendicular distance from the xx-axis to point P(0,0,d)P(0,0,d) is dd. Using Gauss's Law, the electric field is: E=λ2πε0d  ⟹  E∝1dE = \frac{\lambda}{2\pi\varepsilon_0 d} \implies E \propto \frac{1}{d} This corresponds to entry Q. Thus, Q→3\text{Q} \rightarrow \mathbf{3}.

  4. Two infinite parallel wires along (y=0,z=l)(y=0, z=l) with +λ+\lambda and (y=0,z=−l)(y=0, z=-l) with −λ-\lambda (2l≪d2l \ll d): The perpendicular distances from point P(0,0,d)P(0,0,d) to the two wires are (d−l)(d-l) and (d+l)(d+l), respectively. The net electric field along the zz-axis is: E=λ2πε0(d−l)−λ2πε0(d+l)=λ2πε0(2ld2−l2)E = \frac{\lambda}{2\pi\varepsilon_0 (d-l)} - \frac{\lambda}{2\pi\varepsilon_0 (d+l)} = \frac{\lambda}{2\pi\varepsilon_0} \left( \frac{2l}{d^2 - l^2} \right) For l≪dl \ll d: E≈λ(2l)2πε0d2  ⟹  E∝1d2E \approx \frac{\lambda (2l)}{2\pi\varepsilon_0 d^2} \implies E \propto \frac{1}{d^2} This also corresponds to entry R. Thus, R→4\text{R} \rightarrow \mathbf{4}.

  5. An infinite planar sheet of charge in the xyxy-plane with uniform surface charge density σ\sigma: By Gauss's Law, the electric field at a distance dd from an infinite plane sheet of charge is: E=σ2ε0E = \frac{\sigma}{2\varepsilon_0} This is independent of dd. This corresponds to entry P. Thus, P→5\text{P} \rightarrow \mathbf{5}.

Matching Summary:

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

This matching corresponds to Option B.

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