The electric field between two equally but oppositely charged parallel plates
is 2.35 N/C [south]. If a charge of –2.00 C was placed at the midway point
between the plates, what force (magnitude and direction) would act on the
charge? Give the direction as acting north or south.
c) For another set of parallel plates, a charge of –4.00 C experiences a force of
6.00 10–3 N [east]. Determine the magnitude and direction of the electric
field between these plates. Give the direction as east or west.
(continued)
58 G r a d e 1 2 P h y s i c s
Assignment 7.4: Electric Field between the Plates of a Parallel Plate
Capacitor (continued)
d) Label each plate in the diagram below with the appropriate polarity
according to the information from part (c) above
a) To find the force acting on a charge placed at the midway point between the plates, you need to use the equation F = qE, where F is the force, q is the charge, and E is the electric field strength.
Given:
Electric field strength (E) = 2.35 N/C [south]
Charge (q) = -2.00 C
Using the given values in the equation F = qE, we can calculate the force:
F = (-2.00 C) * (2.35 N/C)
F = -4.70 N [south]
Therefore, the magnitude of the force acting on the charge is 4.70 N, and it is directed south.
b) To determine the electric field strength between the plates based on the given force, you can rearrange the equation F = qE to solve for E:
E = F / q
Given:
Force (F) = 6.00 × 10^(-3) N [east]
Charge (q) = -4.00 C
Using the given values in the equation E = F / q, we can calculate the electric field strength:
E = (6.00 × 10^(-3) N) / (-4.00 C)
E = -1.50 × 10^(-3) N/C [east]
Therefore, the magnitude of the electric field strength between these plates is 1.50 × 10^(-3) N/C, and it is directed east.
c) Based on the information from part (b), the electric field between the plates is directed east. Label the plates in the diagram as follows:
Positive plate: West
Negative plate: East