Question

How far apart are two conducting plates that have an electric field strength of 4.5 × 103V/m between them, if their potential difference is 12.5 kV?

Answer 1

Answer:

Explanation:

Given

Electric Field Strength $E=4.5\times 10^{3}\ V/m$

Potential Difference between Plates is given by $V=12.5\ kV$

In conducting plates a Potential difference exist between two plate which accelerate the charge when put between the conducting plates

The potential difference is given by

$\Delta V=Ed$

where E=Electric Field strength

d=distance between Plates

$d=\frac{\Delta V}{E}$

$d=\frac{12.5\times 10^3}{4.5\times 10^{3}}$

$d=2.78\ m$