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Test: Continous Charge Distibution & Gauss Law its Application (October 29) - NEET MCQ


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10 Questions MCQ Test Daily Test for NEET Preparation - Test: Continous Charge Distibution & Gauss Law its Application (October 29)

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Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 1

A metallic spherical shell has an inner radius R1 and outer radius R2. A charge is placed at the centre of the spherical cavity. The surface charge density on the inner surface is

Detailed Solution for Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 1

When a charge +q is placed at the centre of spherical cavity as shown in figure.
Charge induced on the inner surface of shell = - q ... (i)
Charge induced on the outer surface of shell = + q ... (ii)

∴ Surface charge dcnsity on the inner surface = -q/4πR21

Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 2

Match the following and find the correct option.

Detailed Solution for Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 2

Linear charge density, λ = charge/length ; A → q
Surface charge density, σ = charge/area; B → r
Volume charge density, ρ = charge/volume; C → p

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Test: Continous Charge Distibution & Gauss Law its Application (October 29) - 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: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 3

Electric field, E = σ/ε0

Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 4

A uniformly charged conducting sphere of 4.4m diameter has a surface charge density of 60 μC m-2. The charge on the sphere is

Detailed Solution for Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 4

Here D = 2r = 4.4 m,
or r = 2.2 m, σ = 60 μC m-2
Charge on the sphere, q = σ x 4πr2
= 60 x 10-6 x 4 x (22/7) x (2.2)2 = 3.7 x 10-3 C.

Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 5

If  over a surface, then

Detailed Solution for Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 5

As 
where q is charge enclosed by the surface.
when 0, q = 0
i.e., net charge enclosed by the surface must be zero. Therefore, all other charges must be outside the surface. This is because charges outside the surface do not contribute to the electric flux.

Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 6

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: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 6

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

Test: Continous Charge Distibution & Gauss Law its Application (October 29) - 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: Continous Charge Distibution & Gauss Law its Application (October 29) - 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: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 8

A sphere encloses an electric dipole within it. The total flux across the sphere is 

Detailed Solution for Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 8

The net charge enclosed by the sphere is zero.

Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 9

The surface considered for Gauss's law is called

Detailed Solution for Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 9

The surface that we choose for the application of Gauss's law is called Gaussian surface.

Test: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 10

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: Continous Charge Distibution & Gauss Law its Application (October 29) - Question 10

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,

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