Question
What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×104 V1.50\times 10^{4}\textrm{ V}?
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Final Answer

1.50×106 V/m1.50\times 10^{6}\textrm{ V/m}

Solution video

OpenStax College Physics, Chapter 19, Problem 14 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. Two parallel conducting plates are separated by a distance of 1.00 centimeter, which is 1.00 times 10 to the minus 2 meters and the potential difference between them is 1.50 times 10 to the 4 volts. The electric field between two conducting parallel plates is the potential difference divided by the distance by which they are separated. So that's 1.50 times 10 to the 4 volts divided by 1.00 times 10 to the minus 2 meters which is 1.50 times 10 to the 6 volts per meter or you could write units of newtons per coulomb, whichever you prefer, but in the context of this question where we are talking about volts and meters, it makes sense to choose this one as our units.