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Consider a sphere of radius R with charg...

Consider a sphere of radius R with charge density distributed as :
`rho( r) =kr, r le R`
`=0` for `r gt R`
(a) Find the electric field at 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.

Text Solution

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(a) Let us consider a sphere S of radius R and two hypothetic spheres of radius r < R and r > R.
Electric field intensity for point r < R,
`ointvecE.dvecS = 1/epsilon_(0) int rho dN`
But volume, `V = 4/3 pir^(3)`
`dV = 4/3 pi xx 3r^(2) dr`
`dV = 4pir^(2)dr`

`therefore ointvecE.dvecS = 1/epsilon_(0) 4pikint_(0)^(r)r^(3)dr |rho =kr|`
`therefore E=1/(4epsilon_(0)).kr^(2)`
As charge density is positive it means the direction of `vecE` is radially outwards. Electric field intensity for point `r gt R`.
`ointvecE.dvecS = 1/epsilon_(0)intrho dV`
`therefore E=(kR^(4))/(4epsilon_(0)r^(2))`.........(1)
Here, also the charge density is again positive. So, the direction of `vecE` is radially outward.
(b) The two protons must be placed symmetrically on the opposite sides of the centre along a diameter. This can shown by the figure given below.

Charge on sphere,
`q = int_(0)^(R)rho dN = int_(0)^(R)(kr)4pir^(2)dr`
`therefore q=4pik R^(4)/4=2e`
`therefore k = (2e)/(piR^(4))`........(2)
The protons 1 and 2 are embedded at distance r from the centre of the sphere as shown, then attractive force on proton 1, due to charge distribution is,
`F_(1) = eE = (-ekr^(2))/(4epsilon_(0))`.........(3)
And repulsive force on proton 1 due to proton 2 is, Net force on proton 1.
`F =F_(1) + F_(2)`
`therefore F=[(-er^(2))/(4epsilon_(0)).(Ze)/(piR^(4)) + e^(2)/(16 pi epsilon_(0)r^(4))]`
Thus, net force on proton 1 will be zero, when
`therefore (er^(2)2e)/(4epsilon_(0)piR^(4)) = e^(2)/(16piepsilon_(0)r)`
`therefore r=R/(8^(1//4))`
Hence, the protons must be at a distance r from the centre.
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