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.
Gauss's law states that the net electric flux through a closed surface equals
- A.
- B.
- C.
- D. for every closed surface
The electric field in a region is given by . The flux of the field through a rectangular surface parallel to plane is . The area of the surface is _____________ .
An electric field exists in space. A cube of side is placed in the space as per figure given below. The electric flux through the cube is ______ .
An electric field, passes through the surface of area having unit vector . The electric flux for that surface is _________ .
The electric field in a region is given by with . 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 .
The electric field in a region is given by . 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 , and are unit vectors along x, y and z-axes respectively.]
Two point charges and are located at cm and cm, respectively on the -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
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
A point charge causes an electric flux of to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is : (Given )
- A.
- B.
- C.
- D.
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 Ï, and radius R. (I) Ï/ε 0 (B) Electric field at distance r>0 from a uniformly charged infinite plane sheet with surface charge density Ï. (II) Ï/2ε 0 (C) Electric field outside (distance r>0 from center) of a uniformly charged spherical shell with surface charge density Ï, and radius R. (III) 0 (D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density Ï. (IV) 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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