(a) Let us consider a sphere S of radius R and two hypothetic sphere of radius `r lt R and r gt R`
Now, for point rltR, electric field intenstiy will be given by:
`ointE.dS=(1)/(epsi_(0))intrhodV`
[For dV, `V=(4)/(3)pir^(3)impliesdV=3xx(4)/(3)pir^(3)dr=4pir^(2)dr`]
`impliesointE.dS=(1)/epsi_(0))4piKint_(0)^(r)r^(3)dr` (`becausep(r)=Kr`)
`implies(E)4pir^(2)=(4piK)/(epsi_(0))(r^(4))/(4)`
`impliesE=(1)/(epsi_(0))Kr^(2)`
here, charge density of positive.
So, direction of E is radially outwards.
For points `r gtR`, electric field intenstiy will be given by
`ointE.dS=(1)/(epsi_(0))intrho.dV`
`impliesE(4pir^(2))=(4piK)/(epsi_(0))int_(0)^(R)r^(3)dr=(4piK)/(epsi_(0))(R^(4))/(4)`
`impliesE=(K)/(4epsi_(0))(R^(4))/(r^(2))`
Charge density is again positive. So, the directio of E is radially outward.
(b). The two protons must be on the opposite sides of the centre along a diameter folloiwng the rule of symmetry. this can be shown by the figure given below. cahrge on the sphere.
`q=int_(0)^(R)rhodV=int_(0)^(R)(Kr)4pir^(2)dr`
`q=4piK(R^(4))/(4)=2e`
`thereforeK=(2e)/(piR^(4))`
If protons 1 and 2 are embedded at distance r from the centre of the sphere as shown, the attractive force on proton 1 due to charge distribution is
`F_(1)=eE=(-eKr^(2))/(4epsi_(0))`
Repulsive force on proton 1 due to proton 2 is
`F_(2)=(e^(2))/(4piepsi_(0)(2r)^(2))` ltBrgt Net force on proton 1,
`F=F_(1)+F_(2)`
`=F=(-eKr^(2))/(4epsi_(0))+(e^(2))/(16piepsi_(0)r^(2))`
so, `F=[(-er^(2))/(4epsi_(0))(Ze)/(4piR^(4))+(e^(2))/(16piepsi_(0)r^(4))]`
Thus, net force on proton 1 will be zero, when
`(er^(2)2e)/(4epsi_(0)piR^(4))=(e^(2))/(16piepsi_(0)r)`
`impliesr^(4)=(R^(4))/(8)`
`impliesr=(R)/((8)^(1//4))` ltBrgt This is the distance of each of the two protons from the centre of the sphere.