Test: Application of Gauss's Law (NCERT) - NEET MCQ

# Test: Application of Gauss's Law (NCERT) - NEET MCQ

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## 10 Questions MCQ Test NCERTs at Fingertips: Textbooks, Tests & Solutions - Test: Application of Gauss's Law (NCERT)

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Test: Application of Gauss's Law (NCERT) - Question 1

### Two parallel infinite line charges + λ and -λ are placed with a separation distance R in free space. Then etelectric field exactly mid - way between the two line charges is

Detailed Solution for Test: Application of Gauss's Law (NCERT) - Question 1

Electric field at point P due to line charge distribution +λ,
E =  away from +λ

Electric field at point P due to line charge distribution -λ,

E and E have same direction,

Test: Application of Gauss's Law (NCERT) - Question 2

### An electric dipole consists of charges ±2.0 x 10-8C separated by a distance of 2.0 x 10-3 m. It is placed near a long line charge of linear charge density 4.0 x 10-4C m-1 as shown in the figure, such that the negative charge is at a distance of 2.0 cm from the line charge. The force acting on the dipole will be

Detailed Solution for Test: Application of Gauss's Law (NCERT) - Question 2

The electric field at a distance r from the line charge of linear density λ is given by
E = λ/2πε0r
Hence, the field at the negative charge,
E
The force on the negative charge, F1 = (3.6 x 108) (2.0 x 10-8) = 7.2 N towards the line charge
Similarly, the field at the positive charge,
i.e., at r = 0.022 m is E= 3.3 x108 N -1
The force on the positive charge,
F2 = (3.3 x 108) x (2.0 x 10-8) = 6.6 N away from the line charge.
Hence, the net force on the dipole = 7.2N - 6.6N = 0.6 N towards the line charge.

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Test: Application of Gauss's Law (NCERT) - Question 3

### Two infinite plane parallel sheets, separated by a distance d have equal and opposite uniform charge desities σ. Electric field at a point between the sheets is

Detailed Solution for Test: Application of Gauss's Law (NCERT) - Question 3

Electric field, E = σ/ε0

Test: Application of Gauss's Law (NCERT) - Question 4

Two large, thin metal places are parallel and close to each other. On their faces, the plates have surface charge densities of opposite signs and of magnitude 16 x 10-22 C m-2. The electric field between the plates is

Detailed Solution for Test: Application of Gauss's Law (NCERT) - Question 4

Here

Test: Application of Gauss's Law (NCERT) - Question 5

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and magnitude 27 x 10-22C m-2. The electric field  in region II in between the plates is

Detailed Solution for Test: Application of Gauss's Law (NCERT) - Question 5

The value of  in the region II, in between the plates = 3.05 x 10-10N C-1

Test: Application of Gauss's Law (NCERT) - Question 6

A charged ball B hangs from a silk thread S, which makes an angle θ with a large charged conducting sheet P as shown in the figure. The surface charge density of the sheet is proportional to

Detailed Solution for Test: Application of Gauss's Law (NCERT) - Question 6

T sinθ = σq/ε0
T cosθ = mg
∴ tanθ = σq/ε0mg
∴ σ is proportional to tanθ.

Test: Application of Gauss's Law (NCERT) - Question 7

Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell is

Detailed Solution for Test: Application of Gauss's Law (NCERT) - Question 7

For a thin uniformly charged spherical shell, the field points outside the shell at a distance x from the centre is

If the radius of the sphere is R, Q = σ4πR2

This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre.

Test: Application of Gauss's Law (NCERT) - Question 8

A non conducting sphere of radius a has a net charge +q uniformly distributed throughout its volume. A spherical conducting shell having inner and outer radii b and c and net charge −q is concentric with the sphere (see the figure).

(i) The electric field at a distance r from the center of the sphere for r<a is
(ii) The electric field at distance r for A<r<b is 0
(iii) The electric field at distance r for b<r<c is 0
(iv) The charge on the inner surface of the spherical shell is −q
(v) The charge on the outer surface of the spherical shell is +q
Which of the above statements are true?

Test: Application of Gauss's Law (NCERT) - Question 9

There is a solid sphere of radius R having uniformly distributed charge throughout it. What is the relation between electric field E and distance r from the centre (r < R)?

Test: Application of Gauss's Law (NCERT) - Question 10

An early model for an atom considered it to have a positively charged point nucleus of charge Ze, surrounded by a uniform density of negative charge upto a radius R. The atom as a whole is neutral. The electric field at a distance r from the nucleus is (r < R).

Detailed Solution for Test: Application of Gauss's Law (NCERT) - Question 10

Charge on nucleus is =+Ze
total negative charge =−Ze(∵ atoms is electrical neutral)
Negative charge density, ρ= charge/volume =

Consider a Gaussian surface with radius r
By Gauss's theorem

Charge enclosed by Gaussian surface
Using (i)
From (ii),

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