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NEET-UG Biology

Electric flux and Gauss's law with its applications — practice questions

56 questions in the bank on this idea. Below are 10 of them, exactly as they appear in a test.

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  1. Question 1 · 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 xzx-z plane is 6.0Nm2C16.0 \,\mathrm{Nm}^2 \mathrm{C}^{-1}. The area of the surface is _____________ cm2\mathrm{cm}^2.

  2. Question 2 · difficulty L2 · understanding

    An electric field E=(2xi^)NC1\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}.

  3. Question 3 · 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}.

  4. Question 4 · 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 .

  5. Question 5 · 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.]

  6. Question 6 · 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
  7. Question 7 · 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
  8. Question 8 · difficulty L2 · understanding

    A point charge causes an electric flux of 2×104Nm2C1-2 \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=8.85×1012C2 N1 m2\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2} )

    • A. 17.7×108C17.7 \times 10^{-8} \mathrm{C}
    • B. 17.7×108C-17.7 \times 10^{-8} \mathrm{C}
    • C. 15.7×108C15.7 \times 10^{-8} \mathrm{C}
    • D. 15.7×108C-15.7 \times 10^{-8} \mathrm{C}
  9. Question 9 · 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)
  10. Question 10 · difficulty L2 · understanding

    Five charges +q,+5q,2q,+3q+q,+5 q,-2 q,+3 q and 4q-4 q are situated as shown in the figure. The electric flux due to this configuration through the surface SS is :

    • A. qϵ0\frac{q}{\epsilon_0}
    • B. 3qϵ0\frac{3 q}{\epsilon_0}
    • C. 5qϵ0\frac{5 q}{\epsilon_0}
    • D. 4qϵ0\frac{4 q}{\epsilon_0}

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