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Determine the potential difference between two charged parallel plates that are 0.50 cm apart and have an electric field strength of 9.0 V/cm.

Sagot :

chipperrider
[tex]E = \frac{V}{r} \\ V = \frac{E}{r} \\ V = \frac{9.0V/cm}{0.5cm} \\ V = 18V [/tex]
AL2006
If the field strength between the plates is 9 volts per cm, then over a path of 1/2cm parallel to the field, the potential changes by (1/2) x (9 volts) = 4.5 volts.
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