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

Electric field at a distance of 1.45 cm will be
`172.41* 10^4N/C`

Step-by-step explanation:

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

And the potential difference
`V=2.5* 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=(V)/(d)`, here E is electric field, V is potential difference and d is distance

So
`E=(V)/(d)=(2.5* 10^4)/(0.0145)=172.41* 10^4N/C`