ELECTRIC FIELD AND CAPACITORS
When we studied electric fields in the preceding chapter, we noticed that the
electric field created by one or more point source charges varied, depending on the
location. For example, as we move further away from the source charge, the
electric field gets weaker. Even if we stay at the same distance from, say, a single
source charge, the direction of the field changes as we move around. Therefore, we
could never obtain an electric field that was constant in both magnitude and
direction throughout some region of space from point-source charges. However, the
electric field that is created between the plates of a charged parallel-plate
capacitor is constant in both magnitude and direction throughout the region between
the plates; in other words, a charged parallel-plate capacitor can create a uniform
electric field. The electric field, E, always points from the positive plate toward
the negative plate, and its magnitude remains the same at every point between the
plates, whether we choose a point close to the positive plate, closer to the negative
plate, or between them.
Because E is so straightforward (it’s the same everywhere between the plates), the
equation for calculating it is equally straightforward. The strength, E, depends on
the voltage between the plates, V, and their separation distance, d:
V = EdThe equation F = qE showed us that the units of E are N/C. This formula now tells
us that the units of E are also V/m. You’ll see both newtons-per-coulomb and volts-
per-meter used as units for the electric field; it turns out these units are exactly the
same.
Example: The charge on a parallel-plate capacitor is 4 × 10−6 C. If the distance
between the plates is 2 mm and the capacitance is 1uF, what’s the strength of the
electric field between the plates?
Solution: Since Q = CV, we have V = Q/C = (4 × 10−6 C)/(10−6 F) = 4 V. Now,
using the equation V = Ed,
E = V/d = 4 V/(2 · 10−3 m) = 2000 V/mExample: The plates of a parallel-plate capacitor are separated by a distance of 2