A charge Q is distributed uniformly within the material of a hollow sphere of inner and outer radii r_{1} and r_{2} (see the figure). Find the electric field at a point P at a distance x away from the centre for r_{1} < x < r. Draw a rough graph showing the electric field as a function of x for 0 < x < 2r_{2} (see the figure).

#### Solution

Amount of charge present on the hollow sphere = Q

Inner radii of the hollow sphere = r_{1}

Outer radii of the hollow sphere = r_{2}

Consider an imaginary sphere of radius x.

The charge on the sphere can be found by multiplying the volume charge density of the hollow spherical volume with the volume of the imaginary sphere of radius (x-r_{1}).

Charge per unit volume of the hollow sphere,

`ρ = "Q"/( 4/3pi("r"_2^3 -"r"_1^3))`

Charge enclosed by this imaginary sphere of radius x:-

q = ρ × Volume of the part consisting of charge

`"q" = (4/3 pi ("x"^3 -"r"_1^3)"Q")/(4/3pi ("r"_2^3))`

`"q" = (("x"^3 -"r"_1^3))/(("r"_2^3 - "r"_1^3)) "Q"`

According to Gauss's Law,

`oint "E" . "d""s" = "q"/∈_0`

Here, the surface integral is carried out on the sphere of radius x and q is the charge enclosed by this sphere.

`"E" oint "d""s" = "q"/∈_0`

`"E" (4 pi "x"^2) = "q"/∈_0`

` "E" (4 pi "x"^2) = (("x"^3 -"r"_1^3)"Q")/((("r"_2^3 -"r"_1^3)) ∈_0)`

`"E" =("Q"("x"^3 - "r"_1^3))/( 4pi ∈_0 "x"^2 ("r"_2^3 - "r"_1^3)) `

The electric field is directly proportional to x for r_{1} < x < r.

The electric field for r_{2} < x < 2r_{2,}

`"E" = "Q" / (4 pi ∈_0 "x"^2) `

Thus, the graph can be drawn as:-