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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2018 · Paper 2 · Physics · Electrostatics

The electric field E is measured at a point P ( 0,0, d ) ,...

The electric field E is measured at a point P(0,0,d)P ( 0,0, 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 E 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 -I LIST -II
A) E is independent of d P) A point charge at the origin
B) E∝1dE ∝ \dfrac{1}{d} Q) A small dipole with point charges QQ at \left(0,0, l\right)and - Qat\left(0, 0, l\right) . Take2l ≪ d</td></tr><tr><td>C)</td><td></td> </tr> <tr> <td>C)</td> <td>E ∝ \dfrac{1}{{d}^{2}}</td><td>R)</td><td>Aninfinitelinechargecoincidentwiththex–axis,withuniformlinearchargedensity</td> <td>R)</td> <td>An infinite line charge coincident with the x – axis, with uniform linear charge density\lambda</td></tr><tr><td>D)</td><td></td> </tr> <tr> <td>D)</td> <td>E ∝ \dfrac{1}{{d}^{3}}</td><td>S)</td><td>Twoinfinitewirescarryinguniformlinearchargedensityparalleltothex–axis.Theonealong</td> <td>S)</td> <td>Two infinite wires carrying uniform linear charge density parallel to the x – axis. The one along\left(y - 0, z = l\right)hasachargedensityhas a charge density+ \lambdaandtheonealong(y=0,z=−l)and the one along \left(y = 0, z = - l\right) has a charge density –λ– \lambda . Take 2l≪d2l ≪ d
T) Infinite plane charge coincident with the xy – plane with uniform surface charge density

Options

  1. Aa-q,r;b-r,s;c-s,t;d-q,r;
  2. Ba-r;b-t;c-r;d-q,r,s,t;
  3. Ca-t;b-s;c-r,s,t;d-t;
  4. Da-t;b-r;c-p,s;d-q;

Correct answer

D. a-t;b-r;c-p,s;d-q;

Step-by-step solution

(i) Point charge
E=KQd2⇒E∝1d2E = \dfrac{KQ}{{d}^{2}} ⇒ E ∝ \dfrac{1}{{d}^{2}}
(ii)( ii ) Dipole
E=2kpd31+3cos⁡2θE = \dfrac{2kp}{{d}^{3}}\sqrt{1 + 3\cos^{2} \theta}
E∝1d3E ∝ \dfrac{1}{{d}^{3}} for dipole
(iii)( iii ) For line charge
E=2kλdE = \dfrac{2k\lambda}{d}
E∝1dE ∝ \dfrac{1}{d}
(iv) Two line charge
E=2kλd−l−2Kλd+lE = \dfrac{2k\lambda}{d - l} - \dfrac{2K\lambda}{d + l}
=2Kl[d+l−d+ld2−l2]= 2Kl\left[\dfrac{d + l - d + l}{{d}^{2} - {l}^{2}}\right]
E=2Kλ(2l)d2[1−l2d2]E = \dfrac{2K\lambda\left(2l\right)}{{d}^{2}\left[1 - \dfrac{{l}^{2}}{{d}^{2}}\right]}
E∝1d2E ∝ \dfrac{1}{{d}^{2}}
(v) Electric field due to sheet
∈=σ2∈0\in = \dfrac{\sigma}{2{\in}_{0}}
∈=v\in = v is independent of r
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