Electric Field and Electric Potential: Difference between revisions

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== See also ==
== See also ==
===Further reading===





Revision as of 16:05, 25 November 2016

Claimed by Terrence Connors

The Main Idea

In physics, many phenomena that we observe are interrelated in some capacity. In the study of electricity and magnetism, several important physical quantities that play a crucial role in understanding physical interactions are derived from one another. Electric Field is a concept that is discussed early in most Electricity and Magnetism curricula, but it has enormous impact once we discover that it tells us information about Electric Potential, and from that, Potential Energy. This helps physicists to understand both the mechanics of a system, and the quantized nature of a system.

A Mathematical Model

We know that the electric force, given by Coulomb's Law, is [math]\displaystyle{ {\vec{F}=q\vec{E}} }[/math]. We also know that electric field and electric force are closely related, the electric field being equal to the electric force divided by the amount of charge [math]\displaystyle{ {\vec{E}=\frac{\vec{F}}{q}} }[/math].

If we think back to the study of conservation of energy, we know that the change in potential energy of a system is work, which is a force being applied over a distance. Since force and distance are vectors, integrating up over the distance of applied force, we obtain: [math]\displaystyle{ {\Delta U=\int_i^f {\vec{F} • \vec{ds}}} }[/math]. By analogy, we define the electric potential as the energy per coulomb, or potential energy divided by charge: [math]\displaystyle{ {\Delta V=\int_i^f {\vec{E} • \vec{ds}}} }[/math].

Observe both sets of equations: the two for Electric Field and Electric Force, and the two for Electric Potential and Potential Energy. We see that they are all related mathematically. If we integrate the electric force, that is, sum the contributions of force over a finite distance, we obtain the change in potential energy. Dividing by the charge, we obtain the potential difference or electric potential, which we see is simply the integral of the electric field applied over a distance.

A Computational Model

We can model how a system will change in electric potential and potential energy as we move, for example, through a uniform electric field in programs like VPython. One could visualize the electric field, electric force, and quantitatively determine the potential and potential energy as, for instance, a system as simple as a single particle moves through space.

Examples

EXAMPLE 1

The electric field is uniform in this region and equal to < 0, –300, 0> N/C. B is at < 2, 2, 0> m and C is at < 2, 0, 0> m. What is ΔV along a path from B to C?

Solution

In this problem, we are given the electric field and asked to find the change in potential between those two points. The formula that we must apply here is [math]\displaystyle{ \Delta V = -\int_i^f \vec{E} • d\vec{s} }[/math], where the initial point is B and the final point is C, making the distance <2,0,0> m - <2,2,0> m = <0,-2,0> m.

The change in potential therefore is the dot product of the electric field and the change in distance:

-<0,-300,0> N/C • <0,-2,0> m = -600 V

EXAMPLE 2

An HF molecule in the gas phase has an internuclear separation s. We can consider the molecule to be composed of two oppositely charged point charges, H with a positive charge, and F with a negative charge. Calculate the potential difference between points 1 and 2, assuming the distances are much larger than the internuclear separation.

Solution

Here we are asked to find the potential difference between two points on the axis of an electric dipole, as we can determine from the diagram. Since we are told that d>>s, we can approximate the magnitude of the electric field as [math]\displaystyle{ \vec{E}=\frac{1}{4\pi\epsilon_{0}}\frac{2p}{r^3} }[/math].

Therefore, we have:

[math]\displaystyle{ V_2 - V_1 = -\int_1^2 \frac{1}{4\pi\epsilon_{0}}\frac{2p}{x^3} dx = -\frac{1}{4\pi\epsilon_{0}}2p \int_1^2\frac{1}{x^3} dx }[/math]

where 1 and 2 are the distances d_1 and d_2 respectively. Integrating, we obtain:

[math]\displaystyle{ \frac {p}{4\pi\epsilon_{0}}(\frac{1}{d_{2}^2}-\frac{1}{d_{1}^2}) }[/math].

We know that p is equivalent to [math]\displaystyle{ \rho s }[/math], so our final answer is [math]\displaystyle{ V_2 - V_1 = \frac{\rho s}{4\pi\epsilon_{0}}(\frac{1}{d_{2}^2}-\frac{1}{d_{1}^2}) }[/math].

Connectedness

The topic of electric field and electric potential is particularly interesting because it gives scientists and physicists in particular so much insight into the systems they are studying. If we can model the electric field and thus the electric force that an object feels, we gain such a more in-depth understanding of the underlying mechanics of the system, thus enabling us to better model the motion of the system, and understand the energy that the system gains or loses. I think this is incredibly interesting.

There are not many direct impacts of the discussion of electric field and electric potential in the study of Chemical Engineering, but in the study of Chemical Engineering Thermodynamics, this does directly affect chemical potential. Chemical potential is a form of potential energy associated with phase change and chemical reaction. At equilibrium, the chemical potential of a system is zero. The definition of chemical potential is incredibly important in the study of thermodynamics, particularly when analyzing the Gibbs Free Energy of a system.

Today, we are seeing a lot of interesting industrial applications of electric field and electric potential. One that is incredibly interesting is the use of induced electric fields in the separation process of chromatography, which is used to draw compounds out of a mixture by capitalizing on their polarity, solubility, and with the use of electric fields, charge and magnetic affinity.

History

See also

External links

References

Georgia Institute of Technology. PHYS 2212. Intro Physics II. Lecture Notes. Darnton, Nicholas; Greco, Edwin.