The potential between a cone and a plate

[zengodspeed1]

Homework Statement

[/B]
There is a conducting cone with angle α placed so that its vertex is normal to an electrically grounded plate, but electrically insulated from the plate and kept at a constant potential V. Find the potential V and the electric field in the region between the cone and the plate. End effects of the cone are neglected. V is independent of r and φ.

Homework Equations

[/B]
If V is independent of r and Φ we know that the Laplacian reduces to:

we have the boundary condition:

The Attempt at a Solution

Firstly, rearranging the Laplacian to solve for dV:

and integrating with respect to theta gives:

Using the boundary condition for the cone:

I don't really know how I can include the plane into my calculations. Any help would be very much appreciated.

 

[Raihan amin]

What you have calculated is the potential difference between two coaxial cone ,where the inner one has a potential of $V$ at ${\theta}_1$ and the outer one has a potential of 0 at ${\theta}_2$.
So check your calculation again,and find the right equation which fulfill the boundary conditions.Now let ${\theta}_2$ tend to $\frac{\pi}{2}$.Then you will find the desired expression

 

[Raihan amin]

Homework Statement

[/B]
There is a conducting cone with angle α placed so that its vertex is normal to an electrically grounded plate, but electrically insulated from the plate and kept at a constant potential V. Find the potential V and the electric field in the region between the cone and the plate. End effects of the cone are neglected. V is independent of r and φ.

Homework Equations

[/B]
If V is independent of r and Φ we know that the Laplacian reduces to:

View attachment 240191

we have the boundary condition:

View attachment 240192

The Attempt at a Solution

Firstly, rearranging the Laplacian to solve for dV:

View attachment 240193

and integrating with respect to theta gives:

View attachment 240194

Using the boundary condition for the cone:

View attachment 240195

I don't really know how I can include the plane into my calculations. Any help would be very much appreciated.

You can do this by using the boundary condition on $V$,i.e,
$$V(\frac{\pi}{2}) =0$$

 

[zengodspeed1]

You can do this by using the boundary condition on $V$,i.e,
$$V(\frac{\pi}{2}) =0$$

So by doing so I find that:

$$A = \frac{V_{0}}{ln(tan(\frac{\alpha}{2}))}$$

as B reduces to zero.

However I don't really understand what I am now doing with this information.

 

[Raihan amin]

So by doing so I find that:

$$A = \frac{V_{0}}{ln(tan(\frac{\alpha}{2}))}$$

as B reduces to zero.

However I don't really understand what I am now doing with this information.

Now put these value in the Original equation,and you are done.

 

[zengodspeed1]

Now put these value in the Original equation,and you are done.

$$V(\theta) = V_{0} \frac{ln(tan(\frac{\theta}{2}))}{ln(tan(\frac{\alpha}{2}))}$$??

 

[Raihan amin]

Now using this formula,you can find E too.