# The potential between a cone and a plate

• zengodspeed1
In summary,The potential difference between two coaxial cones is found by solving for the potential using the Laplacian and integrating over the cone's surface. The electric field is found by solving for the field using the boundary condition and the potential.
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.

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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

zengodspeed1 said:

## 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,
$$V(\frac{\pi}{2}) =0$$

Raihan amin said:
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.

zengodspeed1 said:
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.

Raihan amin said:
Now put these value in the Original equation,and you are done.
$$V(\theta) = V_{0} \frac{ln(tan(\frac{\theta}{2}))}{ln(tan(\frac{\alpha}{2}))}$$??

Now using this formula,you can find E too.

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## 1. What is the potential between a cone and a plate?

The potential between a cone and a plate refers to the electrical potential difference or voltage that exists between these two objects. It is a measure of the work needed to move an electric charge from the cone to the plate or vice versa.

## 2. How is the potential between a cone and a plate calculated?

The potential between a cone and a plate can be calculated using the formula V = kQ/r, where V is the potential difference, k is the Coulomb's constant, Q is the charge on the cone or plate, and r is the distance between them.

## 3. What factors affect the potential between a cone and a plate?

The potential between a cone and a plate is affected by the distance between them, the charge on each object, and the dielectric constant of the material between them. Additionally, the shape and size of the objects can also impact the potential difference.

## 4. How does the potential between a cone and a plate affect the electric field?

The potential between a cone and a plate is directly related to the electric field between them. The electric field is the gradient of the potential, meaning it is the change in potential over a certain distance. A larger potential difference results in a stronger electric field.

## 5. What applications does the potential between a cone and a plate have?

The potential between a cone and a plate has various applications in the field of electrostatics. It is used in capacitors, where a cone and a plate are used to store electrical energy. It is also used in particle accelerators, where a cone and a plate are used to accelerate charged particles.

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