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 3.22 x 104 V

Answer 1

Answer:

$E=3.22*10^6 N/C$

Explanation:

From the question we are told that:

Separation Distance $d=1.0cm =0.01m$

Potential difference $V=3.22 * 10^4 V$

Generally the equation for Electric Field strength is mathematically given by

 $E=\frac{v}{d}$

 $E=\frac{3.22*10^4}{0.01}$

 $E=3.22*10^6 N/C$