103 Electric Potential in a Uniform Electric Field
103 Electric Potential in a Uniform Electric Field
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Learning Objectives
- Describe the relationship between voltage and electric field.
- Derive an expression for the electric potential and electric field.
- Calculate electric field strength given distance and voltage.
In the previous section, we explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage and electric field. For example, a uniform electric field \(\mathbf{\text{E}}\) is produced by placing a potential difference (or voltage) \(\Delta V\) across two parallel metal plates, labeled A and B. (See (Figure).) Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal a more fundamental relationship between electric potential and electric field. From a physicist’s point of view, either \(\Delta V\) or \(\mathbf{\text{E}}\) can be used to describe any charge distribution. \(\Delta V\) is most closely tied to energy, whereas \(\mathbf{\text{E}}\) is most closely related to force. \(\Delta V\) is a scalar quantity and has no direction, while \(\mathbf{\text{E}}\) is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by \(E\) below.) The relationship between \(\Delta V\) and \(\mathbf{\text{E}}\) is revealed by calculating the work done by the force in moving a charge from point A to point B. But, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiring calculus. We therefore look at a uniform electric field as an interesting special case.
The work done by the electric field in (Figure) to move a positive charge \(q\) from A, the positive plate, higher potential, to B, the negative plate, lower potential, is
The potential difference between points A and B is
Entering this into the expression for work yields
Work is \(W=\text{Fd}\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \); here \(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =1\), since the path is parallel to the field, and so \(W=\text{Fd}\). Since \(F=\text{qE}\), we see that \(W=\text{qEd}\). Substituting this expression for work into the previous equation gives
The charge cancels, and so the voltage between points A and B is seen to be
where \(d\) is the distance from A to B, or the distance between the plates in (Figure). Note that the above equation implies the units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus the following relation among units is valid:
where \(d\) is the distance from A to B, or the distance between the plates.
Dry air will support a maximum electric field strength of about \(3.0×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{V/m}\). Above that value, the field creates enough ionization in the air to make the air a conductor. Th