Eletric potential outside a conductor sphere [closed]

During my chemistry graduation, my Physics III professor gave to us an exercise that would worth an adition in the final mean.

Well, I've finished the graduation, and the exercise was forgoten until now.

So I've came here to ask for help to solve the exercise, that I neve could.

Exercise: Find the potential at the point $P(r,\theta,\phi)$, for a system containing a grounded conductor sphere of radius $R$, and an uniformely charged bar (of constant charge densit $\lambda$ and lenght $L$) in front of the sphere as showed in the image bellow.

It's also stated that $R$ is greather than $L\enspace$ ($R>L$) We were suposed to use the Method of Image Charges, but you can use any method you find apropriate.

EDIT:

After computing all parameters I should be able to substitute on image method formula (1) and find the potential, but I'm still debating if I should integrate all the equation or just the charge distribuition.

$$V = \frac{1}{4 \pi \epsilon_0}\sum_{i=1}^{n} \frac{Q_i}{\Vert\vec{r}-\vec{r_i} \Vert} \tag{1}$$

After doing all the calculations we get:

\begin{align} V = \frac{\lambda}{4 \pi \epsilon_0} \left(\int_{-L/2}^{L/2}\frac{dL}{\Vert\vec{r}-\vec{r_Q} \Vert}\ - \ \int_{-l/2}^{l/2}\frac{dl}{\Vert{}\vec{r}-\vec{r_q} \Vert}\right) \tag{2} \end{align}

where:

\begin{align} \Vert\vec{r}-\vec{r_Q} \Vert &= \sqrt{r^2 + D^2 sec^2{\alpha} - 2rD\ [sin{\theta}\ sin{\phi}\ tg{\alpha} + cos{\theta}]} \tag{3} \\ \Vert\vec{r}-\vec{r_q} \Vert &= \sqrt{r^2 + d^2 sec^2{\beta} - 2rd\ [sin{\theta}\ sin{\phi}\ tg{\beta} + cos{\theta}]} \tag{4} \\ L &= D\ tg{\alpha} \tag{5} \\ l &= d\ tg{\beta} \tag{6} \\ \end{align}

$\alpha$ is the angle formed by the vector that goes from the origin to any point in the charged bar does whit respect to the z axis, similarly $\beta$ is the is the angle formed by the vector that goes from the origin to any point in the image charged bar does whit respect to the z axis.

D, and d, are the distances of the charged bar, and image chagerd bar, respectively, fron the origin of z axis.

$\theta$ and $\phi$ are the common spherical coordinates variables, there trigonometric functions come from the convertion to cartesian coordinates.

If I need to integrate the charge distribution only, or the entire function, just tell me that will solve my physics problem.

closed as off-topic by StephenG, stafusa, John Rennie, Jon Custer, ZeroTheHeroJul 31 '18 at 2:34

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• I would treat the bar as a collection of point charges and find the solution by integrating a known solution for a point charge and a grounded sphere, over the length of the bar. – V.F. Jul 29 '18 at 19:56
• Yes, is the logical path to take, but, I have extreme dificulty in findind a constant to propose a substitution for the integral and simplify it. – liuzp Jul 29 '18 at 22:27