# Electric Field from its Potential of a Half Circle along its Z axis

• AndresPB
In summary, the potential can be calculated using the expression dV = (1/(4*Pi*Epsilon_0))*[λ dl/sqrt(z^2+a^2)]. To find the electric field, you can use ∇ V and label the x and y coordinates differently. Alternatively, you can calculate the potential using V = ∫λadΦ/(4πε0R) and differentiate it. You can also get the electric field on the z-axis from the electric potential V(z) by considering all elements dq on the ring at the same distance from the point of interest and by squeezing all the charge on the ring into a single point charge in the xy plane.
AndresPB
Homework Statement
Find the Electric Potential Energy of half a circle charged with a Linear Charge Density: "Lambda" along its z axis. After that, find the Electric Field "E" deriving its potential energy.
Relevant Equations
Potential Energy dV = (1/(4*Pi*Epsilon_0))*(dQ/|r-r'|)
Electric Field E = - ∇ V

So I figured out the potential is: dV = (1/(4*Pi*Epsilon_0))*[λ dl/sqrt(z^2+a^2)]
.
From that expression: We can figure out that since its half a ring we have to integrate from 0 to pi*a, so we would get:

V = (1/(4*Pi*Epsilon_0))*[λ {pi*a]/sqrt(z^2+a^2)]

In that expression: a = sqrt(x^2+y^2)

What I am having trouble is getting the electric field from this expression. When I do ∇ V, I am getting three components for my electric field in the (x,y,z), directions. But when I analyze the problem. It is clear to me that there is not a "y" component, as there is simetry along this axis. What shall I do?

The answer I am getting at the moment is this ("L" in this case is: (1/(4*Pi*Epsilon_0))*[λ {pi}]):

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It looks like you are using x and y for different things. There is the equation of the semicircle: x2 + y2 = a2. Then there are the coordinates of a point in space (x, y, z), for which you want to find the potential and its derivative. But you can't use the same x and y for both, and try to differentiate by them. You have to label them differently. For example, you could calculate the potential by
V = ∫λadΦ/(4πε0R) where R2 = (x-acosφ)2 + (y-asinφ)2 + z2
Differentiate, and take x = y = 0 along the z axis.

You can get the electric field on the ##z##-axis from the electric potential ##V(z)##. Note that all elements ##dq## on the ring are at the same distance from the point of interest on the ##z##-axis. Would ##V(z)## be any different if you squeezed all the charge on the ring into a single point charge in the ##xy## plane at distance ##a## from the origin ?

## 1. What is an electric field and how is it related to potential?

An electric field is a force field that surrounds charged particles or objects. It is related to potential because the potential at a point in the field is the amount of work needed to move a unit positive charge from infinity to that point against the electric field, per unit charge. In other words, potential is a measure of the strength of the electric field.

## 2. What is a half circle and how does it relate to the electric field?

A half circle is a two-dimensional geometric shape that is formed by taking a complete circle and cutting it into two equal parts. In the context of electric fields, a half circle can be used to model the distribution of charges along a certain path or axis, such as the z-axis. This can help us calculate the electric field at different points along the half circle.

## 3. How is the electric field calculated from the potential of a half circle along its z-axis?

The electric field can be calculated from the potential of a half circle by using the equation E = -∇V, where E is the electric field, ∇ is the gradient operator, and V is the potential. In this case, the potential along the z-axis can be represented by the equation V = λ/2πε0 * ln(r), where λ is the charge density, ε0 is the permittivity of free space, and r is the distance from the center of the half circle to the point of interest.

## 4. How does the electric field vary along the z-axis in a half circle distribution?

The electric field varies along the z-axis in a half circle distribution because the potential varies along this axis, as described by the equation V = λ/2πε0 * ln(r). This means that the electric field will be stronger closer to the center of the half circle and weaker farther away from it. The direction of the electric field will also depend on the sign of the charge density λ.

## 5. What are some real-life applications of the concept of electric field from its potential of a half circle along its z-axis?

The concept of electric field from its potential of a half circle along its z-axis has many real-life applications, such as in the design of electrical circuits and devices. It can also be used in the study of electrostatics, which has applications in a wide range of fields including medicine, engineering, and material science. Additionally, understanding the electric field from its potential can help us better understand and predict the behavior of charged particles and objects in various systems.

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