# Finding voltage from the E field

• iScience
In summary, the conversation is discussing the incorrect use of an equation to find the potential at a point P due to a linear charge density on the x-axis. The equation involves integrating the electric field vectors in the x and z directions, but the integration bounds are unclear and the assumption that the potential is zero far away from the wire is not satisfied. It is suggested to instead use an integral that directly calculates the potential by taking into account the charge density of each small piece of the wire.
iScience
consider..

If I'm trying to find the voltage at point P due to a linear charge density of length L (the white line that lies on the x axis):

then how come it is incorrect to do the following?

$$V=\int{\vec{E_{x}}dx}+\int{\vec{E_{z}}dz}$$?

(the r vector is just to find dE at each point on the charge density of length L)

Last edited:
Hello I,

What exactly does your equation mean ? Ex ? Ez ? what is dx, dz and what are the integration bounds ?

Also: is your endering of the exercise complete ? r ?

$$\vec{E_{x}}$$ and $$\vec{E_{z}}$$ are the electric field vectors in the specified directions, x, and z respectively.

I'm not concerned with specific bounds at this time right now i am only concerned with whether or not i have the right expression for potential at some point in space i decided to call "P".

Also: is your endering of the exercise complete ?

i don't know what you mean

r?

"(the r vector is just to find dE at each point on the charge density of length L)"

You want to find the potential at point P due to the line charge on the x-axis.

If you already, at all points in space, know the electric field, cause by the line charge, then you can integrate, e.g. Ez dz from z =∞ to z = z(P) to get V.

Or Ex dx from x =-∞, z = z(P) to x = 0, z = z(P)

These two under the assumption that V = 0 far away from the wire (which x = ∞, z = z(P) does not satisfy).

They look like your integrals, only one has dx = 0 and the other dz = 0 (makes life easier).

---

My question about the Rendering (sorry) of the original problem was inspired by the fact that I am missing an indication of what z is, what r is, whether the charge density on the line is a constant, ...

---

Since you don't work out your integrals, I am inclined to believe you do not have an expression for ##\vec E(x,z)## for all ##x,z##. You could embark on finding one, but the exercise only asks for the potential.

I take it you do have an expression for V as a function of q and r at hand. So if you chop your line charge into little pieces with length dx, each little piece having a charge ##dq = \lambda dx## and contributing a little ##dV = ... dq = ... dx##, I am sure you can easily write down an integral that finds you V diectly!

Your approach to finding the voltage at point P is not entirely incorrect, but it is not the most accurate method. The equation you have provided is based on the electric field components in the x and z directions, but it does not take into account the electric field in the y direction. In order to accurately calculate the voltage at point P, you need to consider the electric field in all three dimensions.

One way to do this is by using the equation V = -∫E•dl, where E is the electric field vector and dl is the infinitesimal displacement vector along the path of integration. This equation allows you to take into account all three components of the electric field and integrate along the path from the starting point to point P.

Another approach is to use the electric potential due to a continuous charge distribution, given by the equation V = kQ/r, where k is the Coulomb's constant, Q is the total charge, and r is the distance from the point to the charge distribution. This equation takes into account the distance between the point and the charge distribution, rather than integrating along a path.

In summary, while your approach is not entirely incorrect, it is not the most accurate method for calculating the voltage at point P. It is important to consider all three components of the electric field in order to accurately determine the voltage.

## 1. How is voltage related to the electric field?

Voltage is directly proportional to the electric field strength. This means that as the electric field increases, the voltage also increases. Conversely, as the electric field decreases, the voltage decreases.

## 2. Can you explain the equation for finding voltage from the electric field?

The equation for finding voltage from the electric field is V = Ed, where V is the voltage, E is the electric field strength, and d is the distance between the two points. This equation is derived from the definition of voltage, which is the potential energy per unit charge.

## 3. What units are used for voltage and electric field?

Voltage is measured in volts (V) and electric field is measured in volts per meter (V/m). Both units are derived from the SI unit for electric potential, which is the volt.

## 4. How do you determine the direction of the electric field from voltage?

The direction of the electric field is determined by the direction of the voltage gradient. This means that the electric field points in the direction of decreasing voltage. In other words, the electric field points from the positive to the negative electrode.

## 5. Can voltage and electric field exist without each other?

No, voltage and electric field are interdependent. The existence of an electric field is the result of a difference in voltage between two points. Similarly, voltage cannot exist without an electric field to create a potential difference.

• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
25
Views
502
• Introductory Physics Homework Help
Replies
3
Views
380
• Introductory Physics Homework Help
Replies
64
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
3K