Test: Continuous Charge Distribution Gauss Law its Application(1 Nov) - JEE MCQ

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

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

Electric field, E = σ/ε₀

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.

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.

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

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.

According to Gauss's Law, the electric flux through a closed surface depends on the net charge enclosed within the surface. Since an electric dipole consists of two equal and opposite charges, the net charge enclosed by any closed surface around the dipole is zero, resulting in zero electric flux.

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

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

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

E_+λ and E_-λ have same direction,