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Two-point charges, each of magnitude q, are kept at the centre of the cube O and vertex A. Find the flux through the shaded face.

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Important Questions on Electric Charges and Fields

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The figure below shows an imaginary cube of side a. A uniformly charged rod of length a2 moves towards the right at a constant speed v. At t=0, the right end of the rod just touches the left face of the cube. Plot a graph between electric flux ϕE passing through the cube versus time (t).

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The figure shows a cube of side l with edges parallel to the axes. The electric field is in positive y direction but its magnitude changes such that it is 15ε0l2 at the bottom face and 5ε0l2 at the top face in SI units. Find the charge enclosed by the cube.

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Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field is due to

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A long wire of linear charge density λ passes through a cube of side l in such a manner that flux through it is maximum. Now the position of the wire is changed in such a manner that the flux is minimum. The ratio of maximum flux to minimum flux is
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A point charge is placed at a distance a2 perpendicular to the plane and above the centre of a square of side a. The electric flux through the square is
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A sphere of radius R and charge Q is placed inside an imaginary sphere of radius 2R whose centre coincides with the given sphere. The flux related to the imaginary sphere is
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Electric charge is uniformly distributed along a long straight wire of radius 1 mm. The charge per cm length of the wire is Q C cm-1 Coulomb. Another cylindrical surface of radius 50 cm and length 1 m symmetrically encloses the wire as shown in the figure. The total flux passing through the cylindrical surface is

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In the figure shown, a hemispherical bowl of radius R is shown. Electric field of intensity E is present perpendicular to the circular cross section of the hemisphere. The electric flux through the hemisphere is

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