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  • 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 E size 12{E} {} is produced by placing a potential difference (or voltage) Δ V size 12{V} {} across two parallel metal plates, labeled A and B. (See [link] .) 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 Δ V size 12{V} {} or E size 12{E} {} can be used to describe any charge distribution. Δ V size 12{V} {} is most closely tied to energy, whereas E size 12{E} {} is most closely related to force. Δ V size 12{V} {} is a scalar    quantity and has no direction, while E size 12{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 size 12{V} {} below.) The relationship between Δ V size 12{V} {} and E size 12{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 figure shows two vertically oriented parallel plates A and B separated by a distance d. The plate A is positively charged and B is negatively charged. Electric field lines are parallel between the plates and curved at the ends of the plates. A charge q is moved from A to B. The work done W equals q times V sub A B, and the electric field intensity E equals V sub A B over d and potential difference delta V equals q times V sub A B.
The relationship between V size 12{V} {} and E size 12{E} {} for parallel conducting plates is E = V / d size 12{E=V/d} {} . (Note that Δ V = V AB size 12{ΔV=V rSub { size 8{"AB"} } } {} in magnitude. For a charge that is moved from plate A at higher potential to plate B at lower potential, a minus sign needs to be included as follows: –Δ V = V A V B = V AB . See the text for details.)

The work done by the electric field in [link] to move a positive charge q size 12{q} {} from A, the positive plate, higher potential, to B, the negative plate, lower potential, is

W = –Δ PE = q Δ V . size 12{W= - Δ"PE"= - qΔV} {}

The potential difference between points A and B is

–Δ V = ( V B V A ) = V A V B = V AB .

Entering this into the expression for work yields

W = qV AB . size 12{W= ital "qV" rSub { size 8{ ital "AB"} } } {}

Work is W = Fd cos θ size 12{W= ital "Fd""cos"?} {} ; here cos θ = 1 , since the path is parallel to the field, and so W = Fd . Since F = qE , we see that W = qEd . Substituting this expression for work into the previous equation gives

qEd = qV AB . size 12{qEd= ital "qV" rSub { size 8{ ital "AB"} } } {}

The charge cancels, and so the voltage between points A and B is seen to be

V AB = Ed E = V AB d (uniform E - field only),

where d size 12{d} {} is the distance from A to B, or the distance between the plates in [link] . 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:

1 N / C = 1 V / m . size 12{"1 N"/C="1 V"/m} {}

Voltage between points a and b

V AB = Ed E = V AB d (uniform E - field only),

where d size 12{d} {} is the distance from A to B, or the distance between the plates.

What is the highest voltage possible between two plates?

Dry air will support a maximum electric field strength of about 3.0 × 10 6 V/m size 12{3×"10" rSup { size 8{6} } " V/m"} {} . Above that value, the field creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air?


We are given the maximum electric field E size 12{E} {} between the plates and the distance d between them. The equation V AB = Ed size 12{V rSub { size 8{"AB"} } =Ed} {} can thus be used to calculate the maximum voltage.


The potential difference or voltage between the plates is

V AB = Ed . size 12{V rSub { size 8{"AB"} } =Ed} {}

Entering the given values for E size 12{E} {} and d size 12{d} {} gives

V AB = ( 3.0 × 10 6 V/m ) ( 0.025 m ) = 7.5 × 10 4 V


V AB = 75 kV . size 12{V rSub { size 8{"AB"} } ="75" "kV"} {}

(The answer is quoted to only two digits, since the maximum field strength is approximate.)


One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5 cm (1 in.) gap, or 150 kV for a 5 cm spark. This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage will cause a spark if there are points on the surface, since points create greater fields than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a smaller voltage will make a spark jump through humid air. The largest voltages can be built up, say with static electricity, on dry days.

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