Question

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10⁴V?

Answer 1

Given:

$d=\rm 1.00\: cm=0.010 \:m$

$V=1.50×10^4\:\rm V$

The strength of the electric field between two parallel conducting plates

$$E=\frac{V}{d}=\rm \frac{1.50×10^4\:\rm V}{0.010\: m}=1.5*10^6\:\rm V/m$$