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Parallel Plate Capacitor
field between
V = Ed
V = work = Fd = Ed
q q Check these formulas E = F
over then you will do a Check for Understanding
V= Ed
Check for Understanding
Potential difference = 10V
1. Draw the electric field lines between the charged plates
d = 4cm
3. Calculate the electric field intensity between the plates
2. How much work would be done moving 2 electrons across the potential toward the negative plate?
Equipotential Surfaces
Equipotential Surfaces
No work is required to move a charge at a constant speed along an equipotential surface (dotted line)
The electric field at every point along an equipotential surface is perpendicular to the surface
Capacitance is the ratio of charge on either conducting plate to magnitude of the potential difference between two conducting surfaces
Capacitors STORE charge
A parallel plate capacitor has an area of A = 2 X 10­4 m2 and a plate separation of 1 mm.
a) Find its capacitance
b) How much charge is stored on the positive plate if it is connected to a 3V battery?
c) Find the electric field intensity between the plates.
Capacitors STORE electrical energy
HW P. 566 # 21, 25, & 26
21. for 100V d = 0.72m
for 50V d = 1.44m
for 25V d = 2.88m
25. a) C = 1.11 X 10­8F
b) Qmax = 26.55C
26. a) C = 2.43 X 10­6F
b) Q = 2.205 X 10­6 F