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

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

Step-by-step 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=(v)/(d)`

`E=(3.22*10^4)/(0.01)`

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