Lesson 3.4 - Parallel Plates
VIDEO LESSON (29:58 min)
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The electric field intensity (E) between two parallel plates is uniform. (Doesn't depend on position).
The magnitude of the electric field is proportional to the charge density on the plates (charge / area of the plate), i.e.
a) Determine the electric field, \( \vec{E} \)
b) What is the electric field intensity at 1.0 cm from the positive plate.
c) How could you double the field strength?
SOLUTION:
a) What is the electric field intensity between the two plates
b) What is the potential difference at a point 1.0cm from the positive plate
SOLUTION:
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ELECTRIC FIELD NEAR A CONDUCTOR
Recall: Electric Field around Point charge
What happens if we extend the charge into a line (or a plane)?
PARALLEL PLATES
Consider two "stretched" out point charges
The electric field intensity (E) between two parallel plates is uniform. (Doesn't depend on position).
The magnitude of the electric field is proportional to the charge density on the plates (charge / area of the plate), i.e.
where \( \epsilon_o =8.85 \times 10^{-12} \frac{C^2}{N \cdot m^2}=\frac{1}{4 \pi k} \) is the permittivity of free space.
Electric intensity is almost zero outside of the plate (except for fringe effects).
EXAMPLE PROBLEM #1
A pair of conducting plates are mounted 10.0 cm apart. A test charge of \( -3.0 \mu C \) is placed at 5.0 cm from one plate. It experiences a force of \( 450 \mu C \) towards the positive plate.a) Determine the electric field, \( \vec{E} \)
b) What is the electric field intensity at 1.0 cm from the positive plate.
c) How could you double the field strength?
SOLUTION:
POTENTIAL DIFFERENCE AND PARALLEL PLATES
EXAMPLE PROBLEM #2
Two parallel plates are 6.0cm apart and the electric potential between them is 60.0V.a) What is the electric field intensity between the two plates
b) What is the potential difference at a point 1.0cm from the positive plate
SOLUTION:
MILLIKAN OIL DROP APPARATUS
Last modified: Thursday, 19 June 2025, 6:36 PM