A parallel-plate capacitor has a charge Q and plates of area A . What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is E = Q / Aϵ 0 , you might think the force is F = QE = Q 2 / Aϵ 0 . This conclusion is wrong because the field E includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually F = Q 2 /2 Aϵ 0 . Suggestion: Let C = ϵ 0 A / x for an arbitrary plate separation x and note that the work done in separating the two charged plates is W = ∫ F d x .
A parallel-plate capacitor has a charge Q and plates of area A . What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is E = Q / Aϵ 0 , you might think the force is F = QE = Q 2 / Aϵ 0 . This conclusion is wrong because the field E includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually F = Q 2 /2 Aϵ 0 . Suggestion: Let C = ϵ 0 A / x for an arbitrary plate separation x and note that the work done in separating the two charged plates is W = ∫ F d x .
Solution Summary: The author explains how the charge stored on each plate of the capacitor is directly proportional to the potential difference between the plates.
A parallel-plate capacitor has a charge Q and plates of area A. What force acts on one plate to attract it toward the other plate? Because the electric field between the plates is E = Q/Aϵ0, you might think the force is F = QE = Q2/Aϵ0. This conclusion is wrong because the field E includes contributions from both plates, and the field created by the positive plate cannot exert any force on the positive plate. Show that the force exerted on each plate is actually F = Q2/2Aϵ0. Suggestion: Let C = ϵ0A/x for an arbitrary plate separation x and note that the work done in separating the two charged plates is
W
=
∫
F
d
x
.
Example
Two charges, one with +10 μC of charge, and
another with - 7.0 μC of charge are placed in
line with each other and held at a fixed distance
of 0.45 m. Where can you put a 3rd charge of +5
μC, so that the net force on the 3rd charge is
zero?
*
Coulomb's Law Example
Three charges are positioned as seen below. Charge
1 is +2.0 μC and charge 2 is +8.0μC, and charge 3 is -
6.0MC.
What is the magnitude and the direction of the force
on charge 2 due to charges 1 and 3?
93
kq92
F
==
2
r13 = 0.090m
91
r12 = 0.12m
92
Coulomb's Constant: k = 8.99x10+9 Nm²/C²
✓
Make sure to draw a Free Body Diagram as well
Chapter 20 Solutions
Principles of Physics: A Calculus-Based Text, Hybrid (with Enhanced WebAssign Printed Access Card)
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