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

Question

asked Jan 21, 2022 224k views

1 Answer

Zimi · Answer 1 · 2022-01-26T15:37:16+0000

Answer:

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