Verified Solution • 2025–26 Session moderate

QJEEAdv18Phdf58: JEE Advanced 2018 Physics - Class 12 JEEAdv18Phdf58 Solution: The electric field is measured at a point generated due to…

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

LIST - I LIST - II
P. EE is independent of dd 1. A point charge Q at the origin
Q. E ∝ 1/dE\, \propto \,1/d 2. A small dipole with point charges
QQ at (0,0,l)\left( {0,0,l} \right) and −Q-Q at
(0,0,−l).\left( {0,0, - l} \right). Take 2l < < d
R. E ∝ 1/d2E\, \propto \,1/{d^2} 3. An infinite line charge coincident
with the x-axis, with uniform linear charge density λ\lambda
S. E ∝ 1/d3E\, \propto \,1/{d^3} 4. Two infinite wires carrying
uniform linear charge density
parallel to the xx -axis. The one
along (y=0,z=l)\left( {y = 0,z = l} \right) has
a charge density +λ+ \lambda and the one
along (y=0,z=−l)\left( {y = 0,z = - l} \right) has a
charge density Take
5. Infinite plane charge coincident
with the xyxy -plane with uniform surface charge density

Multiple Choice Options

  • A P→5;Q→3,4;R→1;S→2P \to 5;Q \to 3,4;R \to 1;S \to 2
  • B P→5;Q→3;R→1,4;S→2P \to 5;Q \to 3;R \to 1,4;S \to 2
  • C P→4;Q→3;R→1,2;S→4P \to 4;Q \to 3;R \to 1,2;S \to 4
  • D P→4;Q→2,3;R→1;S→5P \to 4;Q \to 2,3;R \to 1;S \to 5

Step-by-Step Verified Solution

For a point charge Q at the origin, we have

E=14πε0Qd2E = {1 \over {4\pi {\varepsilon _0}}}{Q \over {{d^2}}}

⇒E∝1d2\Rightarrow E \propto {1 \over {{d^2}}}

Thus, the correct mapping is R→1R \to 1 .

For a small dipole with point charges Q at (0, 0, l) and −- Q at (0, 0, −- l), 2l << d, we have

E=14πε02Q×2ld3E = {1 \over {4\pi {\varepsilon _0}}}{{2Q \times 2l} \over {{d^3}}}

⇒E∝1d3\Rightarrow E \propto {1 \over {{d^3}}}

Thus, the correct mapping is S→2S \to 2 .

For an infinite line charge coincident with x-axis with uniform linear charge density λ\lambda , we have

E=λ2πε0dE = {\lambda \over {2\pi {\varepsilon _0}d}}

⇒E∝1d\Rightarrow E \propto {1 \over d}

Thus, the correct mapping is Q→3Q \to 3 .

For two infinite wires carrying uniform linear charge density parallel to the x-axis, we have

E=λ2πε0(d−l)−λ2πε0(d+l)=λ2πε0(1d−l−1d+l)E = {\lambda \over {2\pi {\varepsilon _0}(d - l)}} - {\lambda \over {2\pi {\varepsilon _0}(d + l)}} = {\lambda \over {2\pi {\varepsilon _0}}}\left( {{1 \over {d - l}} - {1 \over {d + l}}} \right)

⇒E=λ2πε0(d+l−d+l(d−l)(d+l))=λ2πε0(2l)(d2−l2)\Rightarrow E = {\lambda \over {2\pi {\varepsilon _0}}}\left( {{{d + l - d + l} \over {(d - l)(d + l)}}} \right) = {\lambda \over {2\pi {\varepsilon _0}}}{{(2l)} \over {({d^2} - {l^2})}}

Since 2l &lt; &lt; d

⇒E=λ2πε02ld2\Rightarrow E = {\lambda \over {2\pi {\varepsilon _0}}}{{2l} \over {{d^2}}}

⇒E∝1d2\Rightarrow E \propto {1 \over {{d^2}}}

Thus, the correct mapping is R→4R \to 4 .

For an infinite plane charge coincident with the x-y plane with uniform surface charge density, we have

E=σ2πε0E = {\sigma \over {2\pi {\varepsilon _0}}}

Therefore, E is independent of d.

Thus, the correct mapping is P→5P \to 5 .

Therefore, option (B) is correct.

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