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JEE Advanced Physics Electrostatics 2025 JEE Advanced 2025 (Paper 2)

JEE Advanced Physics Question (2025) — Solution

Question

Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure, the sheets carry uniform surface charge densities which are indicated in terms of _0. The separation between any two consecutive sheets is 1 ~m . The various regions between the sheets are denoted as 1,2,3,4 and 5. If _0=9 C / m ^2, then which of the following statements is/are correct: (Take permittivity of free space _0=9 10^ -12 ~F / m ):

Options

  1. A. In region 4 of the configuration I, the magnitude of the electric field is zero.
  2. B. In region 3 of the configuration II, the magnitude of the electric field is _0 _0 .
  3. C. Potential difference between the first and the last sheets of the configuration I is 5 V .
  4. D. Potential difference between the first and the last sheets of the configuration II is zero.

Answer

A. In region 4 of the configuration I, the magnitude of the electric field is zero.

Step-by-step solution

aligned ( E _4 )_ I = & _0 2 _0 [1-1+1-1-1+1]=0 \\ ( ~V _ First )_ I & = - _0 2 _0 [-1+2-3+4-5] d \\ & = - _0 2 _0 [-3] d = _0 3 ~d 2 _0 aligned aligned & aligned (V_ Last )_I & = - _0 2 _0 [1-2+3-4+5] \\ & = _0 2 _0 [-3 ~d ] aligned \\ & aligned (V_ First -V_ Last )_I & = 3 _0 ~d _0 \\ & = 3 9 10^ -6 1 10^ -6 9 10^ -12 =3 volt aligned aligned aligned & aligned ( E _3 )_ II = & _0 2 _0 [ 1 2 -1+1+1-1+ 1 2 ]= _0 2 _0 \\ & = - _0 2 _0 [-1+2-3+4- 5 2 ] d \\ & = - _0 2 _0 [2-2.5] d = _0 ~d 4 _0 aligned \\ & aligned ( V _ Last )_ II & = - _0 2 _0 [1-2+3-4+ 5 2 ] d \\ & = - _0 2 _0 [6.5-6] d = - _0 ~d 4 _0 aligned \\ & ( ~V _ First - V _ Last )_ II = _0 ~d 2 _0 0 aligned

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Related: Physics — Electrostatics · All PYQ Banks