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Two large parallel conducting plates separated by 10 cm carry equal and opposite surface charge densities such that the electric field between them is uniform. The difference in potential between the plates is 800 V. An electron is released from rest at the negatively charged plate. What is the magnitude of the electric field between the plates?

User Alhassan
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Answer:

Step-by-step explanation:

potential difference between the plates, V = 800 V

Distance between the plates, d = 10 cm = 0.1 m

Electric field between the plates is given by

E = V / d

E = 800 / 0.1 = 8000 N/C

User Ephemera
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