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An infinitely long line charge having linear charge density λ lies at a distance d from center of an imaginary sphere of radius R. Then

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Important Questions on Electrostatics

EASY
The electric field in a region is given by E=25E0i^+35E0j^ with E0=4.0×103 N C-1. The flux of this field through a rectangular surface, area 0.4 m2 parallel to the Y-Z plane is _______ N m2 C-1.
EASY

A charge q is placed at one corner of a cube as shown in figure. The flux of electrostatic field E through the shaded area is:

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HARD
The electric field in a region is given by E=35E0i^+45E0j^ N C-1. The ratio of flux of reported field through the rectangular surface of area 0.2 m2 (parallel to y-z plane) to that of the surface of area 0.3 m2 (parallel to x-z plane) is a:b=a:2, where a=? [Here i^, j^ and k^ are unit vectors along x, y and z -axes respectively]
MEDIUM

Choose the incorrect statement:
(a) The electric lines of force entering into a Gaussian surface provide negative flux.
(b) A charge q is placed at the centre of a cube. The flux through all the faces will be the same.
(c) In a uniform electric field net flux through a closed Gaussian surface containing no net charge, is zero.
(d) When an electric field is parallel to a Gaussian surface, it provides a finite non-zero flux.

Choose the most appropriate answer from the options given below:

MEDIUM
An electric field E=4xi^-y2+1j^ N C-1 passes through the box shown in figure. The flux of the electric field through surfaces ABCD and BCGF are marked as ϕI and ϕII respectively. The value of ϕI-ϕII is (in N m2 C-1) ____________.
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EASY

A cube is placed inside an electric field, E=150y2j^ The side of the cube is 0.5 m and is placed in the field as shown in the given figure. The charge inside the cube is:

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EASY

A point charge of +12 μC is at a distance 6 cm vertically above the centre of a square of side 12 cm as shown in figure. The magnitude of the electric flux through the square will be _________ ×103 N m2 C-1.(Write the value to the nearest integer)

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HARD
An electrical charge distribution is given by volume charge density ρr=αr 0rR when r is the radial coordinate. Find the electric flux due to this distribution through a closed cylindrical surface of radius R enclosing the above charge distribution
HARD
A circular disc of radius R carries surface charge density σ(r)=σ01-rR, where σ0 is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is ϕ0. Electric flux through another spherical surface of radius R4 and concentric with the disc is ϕ. Then the ratio ϕ0ϕ is
EASY

A point charge q is placed at the corner of a cube of side a as shown in the figure. What is the electric flux through the face ABCD ?

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EASY
A point charge of 3.0 μC is at the center of a Gaussian surface of radius 10 cm. What is the net electric flux through the surface?
[use ε0=9×10-12 in S.I. unit]
EASY
The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about 150 N/C, directed inward towards the center of the Earth. This gives the total net surface charge carried by the Earth to be : [Given : O = 8.85 × 1 0 - 1 2   C 2 / N-m 2 ,   R E = 6.37 × 1 0 6 m ]
EASY
Consider the charged cylindrical capacitor. The magnitude of electric field E in its annular region
MEDIUM

A charge q sits at the black corner of a cube as shown in the following figure:

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What is the electric flux through the shaded side?

EASY
The electric field in a region is given as, E=10i^+20j^ V m-1 The net flux passing through a square area of side 2 m, parallel to x-z, plane is,
EASY

The black shapes in the figure below are closed surfaces. The electric field lines are in red. For which case, the net flux through the surfaces is non-zero?

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EASY

A point electric charge Q is placed at a corner of a cube as shown in the figure. What is the electric flux passing through ABCD of the cube? (0 is permittivity of free space)

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HARD
The potential due to an electrostatic charge distribution is Vr=qe-αr4πε0r, where α is positive. The net charge within a sphere centred at the origin and of radius 1/α is
EASY
The electric field in a region of space is given by, E=E0i^+2E0j^ where E0=100 N C-1. The flux of this field through a circular surface of radius 0.02 m parallel to the YZ plane is nearly
EASY
Four closed surfaces and corresponding charge distributions are shown below.

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Let the respective electric fluxes through the surfaces be ϕ1, ϕ2, ϕ3 and ϕ4. Then: