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Physics · Ch 10 — Electrostatics

Practical Way of Calculating Electric Field

10.6.2

Practical Way of Calculating Electric Field

Beyond the fundamental point-charge formula, there is a very practical, everyday way of both calculating AND measuring an electric field, using a pair of charged parallel plates. Consider two parallel plates A and B, separated by a distance d, with a potential difference V applied across them from an external source; the electric field in the region between the plates is UNIFORM, directed from the higher-potential plate (A) to the lower-potential plate (B), as shown in Fig. 10.10.

A positive charge +q placed between the plates experiences a force F due to this uniform field. To move this charge AGAINST the direction of the field -- that is, from the negative plate B toward the positive plate A -- external work must be done on it. If the charge +q is moved the full separation distance d, from B to A, the work done against the field is simply W=Fd.W=Fd. On the other hand, by the very definition of potential difference, the work done in moving a charge q through a potential difference V is W=Vq.W=Vq. Equating these two expressions for the same work done, Vq=Fd⇒Fq=Vd.Vq=Fd\quad\Rightarrow\quad\frac{F}{q}=\frac{V}{d}. But F/qF/q is exactly the definition of electric field E, so this gives the widely-used PRACTICAL formula for the field between parallel plates: E=Vd.E=\frac{V}{d}. This E = V/d relation is, in practice, the most commonly used way of both defining and actually m …

Figure 10.10Fig. 10.10: Electric field between two parallel plates

What this figure shows. Two flat, parallel conducting plates, labelled A and B, held a distance d apart, with plate A shown at the higher (positive) potential and plate B at the lower (negative) potential; a set of straight field-line arrows is drawn running from plate A to plate B in the region between them, all pointing in the same direction (from A to B), consistent with a uniform field. A potential difference V is marked as applied across the two plates from an external source, and the figure sets up the derivation E=V/dE=V/d by considering the work done moving a positive test charg …

Misc Ex.5Example 10.5: Electric field in a spark-plug gap

Worked out. An automobile spark plug has electrodes separated by d=1.25d=1.25 mm, with a potential difference of V=20V=20 V applied across the gap. Using E=V/d=201.25×10−3=1.6×104E=V/d=\frac{20}{1.25\times10^{-3}}=1.6\times10^4 V/m, the resulting electric field is strong enough to ionize the compressed fuel-air mixture in the cylinder and ignite it -- a concrete real-world illustration of how even a modest voltage (20 V) can produce a very large field once the plate separation is made small enough, since …