Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Concepts Of Physics MCQ Edition [Volume 2]PhysicsGauss's Law

A large nonconducting sheet M carries a uniform charge density. Two small, uncharged metal rods A and B are positioned near the sheet, as shown in the figure.

Options

  1. AThe sheet M attracts rod A .
  2. BRod B repels rod A .
  3. CThe sheet M attracts rod B .
  4. DRod A attracts rod B .

Correct answer

D. Rod A attracts rod B .

Step-by-step solution

Let the large nonconducting sheet M have a positive charge density (the reasoning remains the same for a negative charge). The electric field produced by a large (but finite) sheet decreases with distance. This electric field polarizes the uncharged metal rods A and B . For rod A , the end closer to M acquires a negative charge, and the farther end acquires a positive charge. Since the electric field of M is stronger at the closer end, the attractive force on the negative charge is greater than the repulsive force

Practice Gauss's Law on Quantrex Academy →

More from Gauss's Law

In a certain region, the electric field is expressed as E = 3 5 E₀ i + 4 5 E₀ j , where E₀ = 2.0 10^3 N C ⁻¹ . Determine the flux of this field through a rectangular surface havingThe electric field in a certain region is defined by E = E₀ x l i . Determine the charge enclosed within a cubical volume defined by the planes x = 0 , x = a , y = 0 , y = a , z = A charge Q is situated at the centre of a cube. Determine the electric field flux through the six surfaces of the cube.A charge Q is situated at a distance a/2 above the centre of a horizontal square surface of edge a , as shown in the figure. Calculate the flux of the electric field through the sqA charge Q is distributed uniformly over a rod of length l . A hypothetical cube of edge l is positioned such that its centre coincides with one end of the rod. Determine the minimDetermine the net charge in a region in which the electric field is uniform at all points.Determine the flux of the electric field through a spherical surface of radius R caused by a charge of 10⁻⁷ C located at its centre and a second identical charge positioned at a diA charge Q is situated at the centre of an imaginary hemispherical surface. By applying symmetry arguments and Gauss's law, determine the flux of the electric field due to this cha Full Gauss's Law list All Concepts Of Physics MCQ Edition [Volume 2] PYQs