the electric field at a point within the distribution is given by

 E= _o/_o(x/3 - x²/4a)

nudge me if u want me explain the steps

consider the spherical charge distribution. this can be imagined as a number of thin concentric shells of thickness dr. since their thickness is very small the charge density on each shell can be considered uniform and equal to

 = ₀(1-r/a) depending on its radius r.

therefore charge on each shell is given by

dq= Vol  = [4r²dr][_o(1-r/a)]

from gauss law we know that the net electric field due to a charged shell at a pt inside the shell is 0 and at a point outside is equal to the ele f due a pt charge at the centre of that shell.

   from this we can conclude that the ele fld inside the charge dist is only due to the inner shells and not the outer ones

ele field due to each inner shell at a dist x is

 dE= 1/4_odq/x²

dE= 1/4_o * [4r²dr][_o(1-r/a)]/x²

    = (1/_o)_o(r²-r³/a)dr/x²

 

therefore to get the net field integrate it between limits 0 and x

E= _{[0 ]}^{[x ]}  (1/_o)_o(r²-r³/a)dr/x²

upon integrating we get

 

E= _o/_o(r³/3 - r⁴/4a)/x²

sub the limits u'll get

E= _o/_o(x³/3 - x⁴/4a)/x²

   =_o/_o(x/3 - x²/4a)

 

 

at x=a

 

E=_o/_o(a/3 - a²/4a)  = _oa/12_o

 

plz rate me if u like it

 

i got a better methord of solving such problems.

instead of considering multiple shells just consider one gaussian concentric sphere such that it touches the pt at which the ele fld has to be calc.

frm gauss eq

    _o  EdA = q 

  here q   =  _{[ 0]}^[x] dq  =    _[ 0]^[x] _o(r²-r³/a)dr

and,

EdA=EA= E 4x²

sub and solve u get E as the same. the steps r similar to the previous methord but the one is easier to understand.

EdA=EA is apllicable due to spherical symmetry(where A is the surface area of the gaussian sur). all u need to do is to calcutate the total charge enclosed and just plug it in the eq

 

this is applicable to any type of spherical charge dist evn a non conducting sphere.( this is 4 u akku). 

the reason why volume is 4r²dr is because its the volume of the shell.

volume of shell= vol of sphere of radius r+dr - vol of sphere of rad r

                     = 4/3(r+dr)³- 4/3r³

                     =4/3( r³+ dr³ + 3rdr( r+ dr)- r³)

                     =4/3(dr³ + 3r²dr + 3rdr²)

now dr² and dr³ are negligible so we are left with

             vol= 4/3*3r²dr

                 = 4r²dr