![]() ![]() This limits the voltages that can exist between conductors, perhaps on a power transmission line. 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. (The answer is quoted to only two digits, since the maximum field strength is approximate.) The potential difference or voltage between the plates is The equation can thus be used to calculate the maximum voltage. We are given the maximum electric field between the plates and the distance between them. Then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air? This allows a discharge or spark that reduces the field. Above that value, the field creates enough ionization in the air to make the air a conductor. Per coulomb thus the following relation among units is valid:Įxample 19.4 What Is the Highest Voltage Possible between Two Plates?ĭry air will support a maximum electric field strength of about. We already know the units for electric field are newtons Note that the above equation implies the units for electric field are volts per meter. Where is the distance from A to B, or the distance between the plates in Figure 19.5. The charge cancels, and so the voltage between points A and B is seen to be Substituting this expression for work into the previous equation gives Work is here, since the path is parallel to the field, and so. The potential difference between points A and B isĮntering this into the expression for work yields The work done by the electric field in Figure 19.5 to move a positive charge from A, the positive plate, higher potential, to B, the negative plate, lower potential, is For a charge that is moved from plate A at higher potential to plate B at lower potential, a minus sign ![]() We therefore look at a uniform electric field as an interesting special case.įigure 19.5 The relationship between and for parallel conducting plates is. But, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitraryĬharge distributions, requiring calculus. The relationship between and is revealed by calculating the work done by the force in moving a charge from point A to point B. (Note that the magnitude of the electric field strength, a scalar quantity, is represented by below.) is a scalar quantity and has no direction, while is a vector quantity, having both is most closely tied to energy, whereas is most closely related to force. From a physicist’s point of view, either orĬan be used to describe any charge distribution. (See Figure 19.5.)Įxamining 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. (or voltage) across two parallel metal plates, labeled A and B. For example, a uniform electric field is produced by placing a potential difference In this section, we will explore the relationship between voltage and electric field. In the previous section, we explored the relationship between voltage and energy. ![]()
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