Question

Question Part Points Submissions Used The difference in potential between the accelerating plates of a TV set is about 218 V. If the distance between these plates is 1.29 cm, find the magnitude of the uniform electric field (N/C) in this region. Consider entering your answer using scientific notation.

Answer 1

`1.69\cdot 10^4 N/C`

Step-by-step explanation:

The relationship between electric field strength and potential difference is:

`E=(V)/(d)`

where

E is the electric field strength

V is the potential difference

d is the distance between the plates

Here we have

V = 218 V

d = 1.29 cm = 0.0129 m

So, the electric field is

`E=(218 V)/(0.0129 m)=16900 N/C = 1.69\cdot 10^4 N/C`