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Consider a sphere of radius R with harge density distributed as p(r)=kr fro `r le R=0` for `r gt R`.
(a). Find the electric field as all points r.
(b) Suppose the total charge on the sphere is 2e where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.

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image
(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.
image
`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.

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