# Method of Images from an earthed half cylinder and plane

• JoelKTH
In summary, the figure attached shows that the electric field has a potential at the top of the half cylinder, which is the same as the potential at the bottom.
JoelKTH
Homework Statement
Given a linecharge ##p_l## straight across a half-cylinder on a large ground plane, as shown in figure. Determine the magnitude and direction of the force on the line charge per unit length. Also verify that the dimension of the expression is correct. The half-cylinder and the ground plane are both metal and are treated as perfectly conducting
Relevant Equations
To get the Electric field ##d\mathbf{E} = \frac{dq'}{4\pi\epsilon_0}\frac{\mathbf{r} - \mathbf{r}'}{\left|\mathbf{r} - \mathbf{r}'\right|^3}##

Lorentz law ##\mathbf{F} = q\mathbf{E}##
Hi,

I want to solve the problem by method of mirroring and by using the electric field by doing superposition and then adding them up to use in Lorentz law to get the force. I have attached a figure that represents the problem.

How do I know from the figure that $-p_l$ is from the half-cylinder and why is it on the positive part of z-axis? The mirroring have opposite direction of $p_l$ depending if its from the linecharge or from the metal half cylinder.

How do I know that the ##-p_l## is on the distance ##z=\frac{a^2}{h}##?

How should I pick my source and field vectors? I believe I should have my field vector as ##r = h \hat{\mathbf{z}}##.

Kind regards,

#### Attachments

• mirror.jpg
24 KB · Views: 78
Your first figure shows the line charge as a short horizontal line of unstated length, terminating above the middle of the half cylinder. Your second looks like it is an infinite line both ways.
Can you calculate the potential at the top of the half cylinder?

Last edited:
haruspex said:
Your first figure shows the line charge as a short horizontal line of unstated length, terminating above the middle of the half cylinder. Your second looks like it is an infinite line both ways.
Can you calculate the potential at the top of the half cylinder?
I actually solved it by knowing that V=0 both at the top of the cylinder and the bottom. I used two equations by adding R to the figure.

#### Attachments

• equation mirror.jpg
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• figur.jpg
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I managed to get the same answer as my professor, however we defined the charges differently.
Am I free to choose both alternative A and B(see figure attached) or is one of them to prefer?

#### Attachments

• mirror charge.jpg
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Last edited:
I was a bit confused when to use what equation. From my calculations i used ##E =\frac{\rho_l}{2\pi\epsilon_0} \frac{(r-r')}{|r-r'|^3}## and got the correct result. I don't quite get why this equation is to be used compared to my first instinct:

##dE =\frac{dq}{4\pi\epsilon_0} \frac{(r-r')}{|r-r'|^3}##,
##dq= \rho_l dr## --> ##E= \frac{\rho_l}{4\pi\epsilon_0} \int \frac{(r-r')}{|r-r'|^3} dr##

How can I know which one to use?

Here are my calculations and answer:

#### Attachments

• calculations.jpg
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• answer.jpg
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Then I did dimensionanalysis and its ok. But I noticed that one needs to check the boundary conditions.
My professor writes in the solution:
To convince ourselves that the charge distribution is correct, lets examine the boundary condition. First, let z = 0. The symmetry in the choice of the line charge implies that at ##r = x\hat{x}##, we obtain (see attached figure). That means z=0 is normal to the Electric field on the plane.

I am a bit lost what is happening here. Because now my professor has used the formula that I thought I was not supposed to use.
He continues to use it to examine the semicircle: we note that on the surface of the cylinder, we have r = ##\hat{x}a \cos \theta + \hat{z}a \sin \theta##(see second attached figure).
Then, since ##\hat{x}\cos\theta + \hat{z}\sin\theta## is normal to the cylinder (parallel to r), the electric field satisfies the normal condition to the surface, and E is the unique solution.

Making me very confused.

#### Attachments

• E-fält kontroll .png
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• E-fält kontroll 2.png
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JoelKTH said:
I managed to get the same answer as my professor, however we defined the charges differently.
Am I free to choose both alternative A and B(see figure attached) or is one of them to prefer?
A is not going to give a zero potential over the top of the dome.

## What is the Method of Images in electrostatics?

The Method of Images is a mathematical technique used in electrostatics to simplify the calculation of electric fields and potentials. It involves replacing the actual problem with an equivalent one that has the same boundary conditions but is easier to solve. This is done by introducing imaginary charges (image charges) that replicate the effect of the boundaries, such as conductors or grounded surfaces.

## How does the Method of Images apply to an earthed half-cylinder and plane configuration?

In the case of an earthed half-cylinder and plane, the Method of Images involves placing image charges in such a way that the potential on the surface of the half-cylinder and the plane is zero. This typically involves placing image charges symmetrically relative to the plane and the cylindrical surface, ensuring that the boundary conditions are satisfied.

## What are the boundary conditions for the earthed half-cylinder and plane problem?

The boundary conditions for this problem are that the electric potential on the surface of the earthed half-cylinder and the plane must be zero. This means that any solution to the problem must ensure that the potential function satisfies these conditions on the specified surfaces.

## How do you determine the position and magnitude of image charges for this configuration?

The position and magnitude of image charges are determined by the geometry of the problem and the requirement to satisfy the boundary conditions. For an earthed half-cylinder and plane, image charges are typically placed at symmetrical positions relative to the cylinder and plane. The magnitudes are chosen so that the combined potential from the real and image charges results in zero potential on the earthed surfaces.

## Can you provide a specific example of using the Method of Images for an earthed half-cylinder and plane?

Consider a point charge +q located at a distance d from the axis of an earthed half-cylinder of radius R, with the cylinder's flat side lying on the earthed plane. To use the Method of Images, you would place an image charge -q at a symmetrical position inside the cylinder, ensuring that the potential on the cylinder's surface is zero. Additionally, you would place another image charge -q at a symmetrical position relative to the plane to ensure the potential on the plane is zero. The exact positions and magnitudes of these image charges can be calculated based on the geometry and distances involved.

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