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

Step-by-step explanation:

Given

Electric Field Strength
`E=4.5* 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=(\Delta V)/(E)`

`d=(12.5* 10^3)/(4.5* 10^(3))`

`d=2.78\ m`