separated by 8.50 cm, is 8.00 × 10

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Answer a and b only.. no handwritten..

The electric field strength between two parallel conducting plates, separated by 8.50 cm, is 8.00 × 10¹
V
m
a) What is the potential difference between the plates?
V=
V
b) The plate with the lowest potential is taken to be zero volts. What is the potential 1.00 cm from that
plate?
V-
c) What is the potential 3.00 cm from the plate at the higher potential?
V=
V
Transcribed Image Text:The electric field strength between two parallel conducting plates, separated by 8.50 cm, is 8.00 × 10¹ V m a) What is the potential difference between the plates? V= V b) The plate with the lowest potential is taken to be zero volts. What is the potential 1.00 cm from that plate? V- c) What is the potential 3.00 cm from the plate at the higher potential? V= V
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