Q16PE.
Question
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 \(15.0kV\)?
Step-by-Step Solution
Verified Answer
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.
1Principle
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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