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NEET-UG and JEE Main Physics · Class 12

Electric flux and Gauss's law with its applications — NEET & JEE Questions

Electric flux and Gauss's law with its applications is a topic in Electrostatics, a Class 12 Physics chapter of the NEET-UG and JEE Main syllabus. TestBudy holds 42 published NEET-UG and JEE Main questions on Electric flux and Gauss's law with its applications, each with the marking scheme of the real paper.

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Electric flux and Gauss's law with its applications MCQs with solutions

10 of the 42 published questions on this topic, exactly as they appear in a test. Answers and worked solutions come with your result rather than on this page.

  1. Question 1 · difficulty L1 · recall

    Gauss's law states that the net electric flux through a closed surface equals

    • A. Qenc/ϵ0Q_{enc}/\epsilon_0
    • B. ϵ0Qenc\epsilon_0 Q_{enc}
    • C. Qenc2/ϵ0Q_{enc}^2/\epsilon_0
    • D. 00 for every closed surface
  2. Question 2 · difficulty L2 · understanding

    The electric field in a region is given by E→=(2i^+4j^+6k^)×103 N/C\overrightarrow{\mathrm{E}}=(2 \hat{i}+4 \hat{j}+6 \hat{k}) \times 10^3 \mathrm{~N} / \mathrm{C}. The flux of the field through a rectangular surface parallel to x−zx-z plane is 6.0 Nm2C−16.0 \,\mathrm{Nm}^2 \mathrm{C}^{-1}. The area of the surface is _____________ cm2\mathrm{cm}^2.

  3. Question 3 · difficulty L2 · understanding

    An electric field E⃗=(2xi^)NC−1\vec{E}=(2 x \hat{i}) N C^{-1} exists in space. A cube of side 2 m2 \mathrm{~m} is placed in the space as per figure given below. The electric flux through the cube is ______ Nm2/C\mathrm{Nm}^2 / \mathrm{C}.

  4. Question 4 · difficulty L2 · understanding

    An electric field, E→=2i^+6j^+8k^6\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}} passes through the surface of 4 m24 \mathrm{~m}^2 area having unit vector n^=(2i^+j^+k^6)\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right). The electric flux for that surface is _________ Vm\mathrm{Vm}.

  5. Question 5 · difficulty L2 · understanding

    The electric field in a region is given by E→=25E0i^+35E0j^\overrightarrow E = {2 \over 5}{E_0}\widehat i + {3 \over 5}{E_0}\widehat j with E0=4.0×103NC{E_0} = 4.0 \times {10^3}{N \over C}. The flux of this field through a rectangular surface area 0.4 m 2 parallel to the Y-Z plane is __________ Nm 2 C −-1 .

  6. Question 6 · difficulty L2 · understanding

    The electric field in a region is given by E→=(35E0i^+45E0j^)NC\overrightarrow E = \left( {{3 \over 5}{E_0}\widehat i + {4 \over 5}{E_0}\widehat j} \right){N \over C}. The ratio of flux of reported field through the rectangular surface of area 0.2 m 2 (parallel to y −- z plane) to that of the surface of area 0.3 m 2 (parallel to x −- z plane) is a : b, where a = __________ [Here i^{\widehat i}, j^{\widehat j} and k^{\widehat k} are unit vectors along x, y and z-axes respectively.]

  7. Question 7 · difficulty L2 · understanding

    Two point charges 8 μC8\ \mu C and −2 μC-2\ \mu C are located at x=2x = 2 cm and x=4x = 4 cm, respectively on the xx-axis. The ratio of electric flux due to these charges through two spheres of radii 3 cm and 5 cm with their centers at the origin is ________.

    • A. 4 : 1
    • B. 3 : 4
    • C. 4 : 3
    • D. 4 : 5
  8. Question 8 · difficulty L2 · understanding

    Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The outer body of an aircraft is made of metal which protects persons sitting inside from lightning strikes. Reason (R) : The electric field inside the cavity enclosed by a conductor is zero. In the light of the above statements, choose the most appropriate answer from the options given below :

    • A. Both (A) and (R) are correct but (R) is not the correct explanation of (A)
    • B. (A) is correct but (R) is not correct
    • C. Both (A) and (R) are correct and (R) is the correct explanation of (A)
    • D. (A) is not correct but (R) is correct
  9. Question 9 · difficulty L2 · understanding

    A point charge causes an electric flux of −2×104Nm2C−1-2 \times 10^4 \mathrm{Nm}^2 \mathrm{C}^{-1} to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is : (Given ϵ0=8.85×10−12C2 N−1 m−2\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2} )

    • A. 17.7×10−8C17.7 \times 10^{-8} \mathrm{C}
    • B. −17.7×10−8C-17.7 \times 10^{-8} \mathrm{C}
    • C. 15.7×10−8C15.7 \times 10^{-8} \mathrm{C}
    • D. −15.7×10−8C-15.7 \times 10^{-8} \mathrm{C}
  10. Question 10 · difficulty L2 · understanding

    Match List - I with List - II. List - I List - II (A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (I) σ/ε 0 (B) Electric field at distance r>0 from a uniformly charged infinite plane sheet with surface charge density σ. (II) σ/2ε 0 (C) Electric field outside (distance r>0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (III) 0 (D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ. (IV) σϵ0r2\frac{\sigma}{\epsilon_0 r^2} Choose the correct answer from the options given below :

    • A. (A)-(IV), (B)-(I), (C)-(III), (D)-(II)
    • B. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
    • C. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
    • D. (A)-(IV), (B)-(II), (C)-(III), (D)-(I)

Common mistakes in Electric flux and Gauss's law with its applications

Recorded against questions in this topic that are not shown above, so none of these points at an answer you can see on this page.

  • Confuses electric-field magnitude with total flux.
  • Includes external charges in the net-flux calculation.
  • Equates zero net flux with zero electric field at every point.

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