Consider a
parallel-plate capacitor
· Initially the plates are separated by a vacuum and connected
to a battery, giving the charge on the plates +Q and
–Q
· The battery is now removed and the charge on the
plates remains constant
· The electric field between the plates is uniform and has a
magnitude of E0
· Meanwhile the separation between plates is d
|
· When a dielectric is placed in the electric field between
the plates, the molecules of the dielectric tend to become oriented with their
positive ends pointing toward the negatively charged plate and vice versa
· The result is a buildup of positive
charge on one surface of the dielectric and of negative charge on the other
|
· The number of field
lines within the dielectric is reduced thus the applied
electric field E0 is partially
canceled
· The new
electric field, E is electric field between the plates is uniform and has a
magnitude of E0
· The new electric field strength (E < E0)
is less, then the potential difference, V across
the plates is less
V
= Ed
· Since V is smaller while Q remains
the same the capacitance
C
= Q/V
is increased by
the dielectric
|
Thursday, December 28, 2017
How dielectric can increase the capacitance of a capacitor?
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