Question

A computer monitor accelerates electrons and directs them to the screen in order to create an image. If the accelerating plates are 1.45 cm apart, and have a potential difference of 2.50 x 10^4 V , what is the magnitude of the uniform electric field between them?

Answer 1

Answer:

Electric field at a distance of 1.45 cm will be $172.41\times 10^4N/C$

Explanation:

We have given the distance d = 1.45 cm = 0.0145 m

And the potential difference $V=2.5\times 10^4volt$

There is a relation between potential difference and electric field

Electric field at a distance d due to a potential difference is given by

$E=\frac{V}{d}$, here E is electric field, V is potential difference and d is distance

So $E=\frac{V}{d}=\frac{2.5\times 10^4}{0.0145}=172.41\times 10^4N/C$