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

Answer:

$1.69\cdot 10^4 N/C$

Explanation:

The relationship between electric field strength and potential difference is:

$E=\frac{V}{d}$

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

$E=\frac{218 V}{0.0129 m}=16900 N/C = 1.69\cdot 10^4 N/C$