# Question 84c22

Apr 20, 2017

I am going to assume you know something of Gauss' Law and what a Gaussian surface is.

Gauss's Law tells us that:

$\int {\int}_{S} m a t h b f E \cdot d m a t h b f S = \frac{\sum {Q}_{\text{enclosed}}}{\varepsilon}$, 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 $H$.

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:

$\int {\int}_{S} m a t h b f E \left(\rho\right) \cdot d m a t h b f S = E \left(\rho\right) 2 \pi \rho H q \quad \triangle$

We can also use your density function, ${\rho}_{v} = \frac{10}{\rho} \left[p \frac{C}{m} ^ 3\right]$, to make a volume integral:

 (sum Q_("enclosed"))/(varepsilon) = 1/epsilon int_V ρ_v(rho) \ color(blue)(dV)#

$= \frac{{10}^{- 11}}{\epsilon} {\int}_{\theta = 0}^{2 \pi} \setminus {\int}_{z = 0}^{H} \setminus {\int}_{\rho = o}^{\rho} \frac{1}{\cancel{\rho}} \setminus \textcolor{b l u e}{\cancel{\rho} \setminus d \rho \setminus d \theta \setminus d z}$

Due to symmetry, this simplifies to:

$\frac{\sum {Q}_{\text{enclosed}}}{\varepsilon} = 2 \pi H \frac{{10}^{- 11}}{\epsilon} \int \setminus d \rho$

$= 2 \pi H \frac{{10}^{- 11}}{\epsilon} \left(\rho + {c}_{1}\right) q \quad \square$

Because $\triangle = \square$:

$E \left(\rho\right) 2 \pi \rho H = 2 \pi H \frac{{10}^{- 11}}{\epsilon} \left(\rho + {c}_{1}\right)$

$\implies m a t h b f E \left(\rho\right) = \frac{{10}^{- 11}}{\epsilon} \left(1 + {c}_{1} / \rho\right) \quad m a t h b f {e}_{\rho} q \quad \star$

That's our electric field. We now call on this relationship between potential $V$ and electric field $m a t h b f E = - \nabla V$, which simplifies, as $\rho$ is the only practical independent variable, to: $m a t h b f E = - \frac{\partial V}{\partial \rho}$.

We integrate $\star$ to get potential function $V \left(\rho\right)$:

$V \left(\rho\right) = - \frac{{10}^{- 11}}{\epsilon} \left(\rho + {c}_{1} \ln \rho + {c}_{2}\right)$

We have 2 unknowns but also 2 points, ie: $V \left(2\right) = 0 , V \left(5\right) = 60$.

The first point tells us that:

$0 = 2 + {c}_{1} \ln 2 + {c}_{2} \implies {c}_{2} = - 2 - {c}_{1} \ln 2$

$\implies V \left(\rho\right) = \frac{{10}^{- 11}}{\epsilon} \left(\left(\rho - 2\right) + {c}_{1} \ln \left(\frac{\rho}{2}\right)\right)$

The second point tells us that:

$60 = - \frac{{10}^{- 11}}{\epsilon} \left(3 + {c}_{1} \ln \left(\frac{5}{2}\right)\right)$

$\implies {c}_{1} = \frac{- \frac{60 \epsilon}{{10}^{- 11}} - 3}{\ln \left(\frac{5}{2}\right)}$

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 $\star$:

$\implies m a t h b f E \left(\rho\right) = \frac{{10}^{- 11}}{\epsilon} \left(1 + \frac{\frac{- \frac{60 \epsilon}{{10}^{- 11}} - 3}{\ln \left(\frac{5}{2}\right)}}{\rho}\right) \quad m a t h b f {e}_{\rho}$

$= \frac{{10}^{- 11}}{8.854 \times {10}^{- 12} \cdot 3.6} \left(1 + \frac{\frac{\frac{- 60 \left(8.854 \times {10}^{- 12} \cdot 3.6\right)}{{10}^{- 11}} - 3}{\ln \left(\frac{5}{2}\right)}}{\rho}\right) \quad m a t h b f {e}_{\rho}$

$= \left(0.314 - \frac{66.51}{\rho}\right) \quad m a t h b f {e}_{\rho}$