How far apart are two conducting plates that have an electric field strength of 4.50×103 V/m4.50\times 10^{3}\textrm{ V/m} between them, if their potential difference is 15.0 kV?
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Final Answer

3.33 m3.33\textrm{ m}

Solution video

OpenStax College Physics, Chapter 19, Problem 16 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. Two parallel conducting plates have an electric field between them of 4.50 times 10 to the 3 volts per meter and a potential difference of 15.0 kilovolts, which is 15.0 times 10 to the 3 volts. Now electric field between two conducting parallel plates is the potential difference divided by the distance by which they are separated and we can solve for this separation d by multiplying both sides by d over E and so the separation is voltage divided by electric field and this works out to 3.33 meters.