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How far apart are two conducting plates that have an electric field strength of \(4.50 \times {10^3}\;V/m\) between them, if their potential difference is \(15.0kV\)?

Short Answer

Expert verified

The two conducting plates that have an electric field strength of \(4.50 \times {10^3}\;V/m\) between them is \(3.33\;m\)apart.

Step by step solution

01

Principle

The potential difference between two points separated by a distance\(d\)in a homogeneous electric field of magnitude\(E\)is\(\Delta V = Ed{\rm{ }}......(1)\).

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