I am going to assume you know something of Gauss' Law and what a Gaussian surface is.
Gauss's Law tells us that:
int int_S mathbf E cdot d mathbf S = (sum Q_("enclosed"))/(varepsilon), where varepsilon = varepsilon_r varepsilon_o
We centre a cylindrical Gaussian surface at rho = 0, so that it's length runs along the cylindrical z-axis. We give our particular surface a notional height H.
![Van Gogh]()
Due to symmetry, the E-field will point directly away from the surface of the cylinder at all points.
We can now use the cylindrical symmetry to say that:
int int_S mathbf E(rho) cdot d mathbf S = E(rho) 2 pi rho H qquad triangle
We can also use your density function, rho_v=10/rho [pC/m^3], to make a volume integral:
(sum Q_("enclosed"))/(varepsilon) = 1/epsilon int_V ρ_v(rho) \ color(blue)(dV)
=(10^(-11) )/epsilon int_(theta = 0)^(2 pi) \ int_(z =0)^H \ int_(rho = o)^rho 1/cancel(rho) \ color(blue) (cancel(rho) \ d rho \ d theta \ d z)
Due to symmetry, this simplifies to:
(sum Q_("enclosed"))/(varepsilon) = 2 pi H (10^(-11) )/epsilon int \ d rho
= 2 pi H (10^(-11) )/epsilon ( rho + c_1) qquad square
Because triangle = square:
E(rho) 2 pi rho H = 2 pi H (10^(-11) )/epsilon ( rho + c_1)
implies mathbf E(rho) = (10^(-11) )/epsilon ( 1 + c_1/rho) quad mathbf e_(rho) qquad star
That's our electric field. We now call on this relationship between potential V and electric field mathbf E = - nabla V, which simplifies, as rho is the only practical independent variable, to: mathbf E = - (partial V)/(partial rho).
We integrate star to get potential function V(rho):
V(rho) = -(10^(-11) )/epsilon ( rho + c_1 ln rho + c_2)
We have 2 unknowns but also 2 points, ie: V(2)= 0, V(5) =60.
The first point tells us that:
0 = 2 + c_1 ln 2 + c_2 implies c_2 = - 2 - c_1 ln 2
implies V(rho) = (10^(-11) )/epsilon ( (rho-2) + c_1 ln (rho/2) )
The second point tells us that:
60 = -(10^(-11) )/epsilon ( 3 + c_1 ln (5/2) )
implies c_1 = ( -(60 epsilon)/ (10^(-11) ) - 3 )/(ln (5/2))
Rounding seems to have a material effect on the result you have posted, so what follows so this is not pretty arithmetic-wise. From here, you might want to nail some of the numbers first. This is presented in a very ugly way, but it sorta gets us an answer that is uncannily like the one you want.
From star:
implies mathbf E(rho) = (10^(-11) )/epsilon ( 1 + ( (- (60 epsilon)/ (10^(-11) ) - 3 )/(ln (5/2)) )/rho) quad mathbf e_(rho)
= (10^(-11) )/(8.854 xx 10^(-12) cdot 3.6) ( 1 + ( ( (-60 (8.854 xx 10^(-12) cdot 3.6))/ (10^(-11) ) - 3 )/(ln (5/2)) )/rho) quad mathbf e_(rho)
=( 0.314 - 66.51/rho ) quad mathbf e_(rho)
