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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?

Respuesta :

Answer:

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

Explanation:

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

And the potential difference [tex]V=2.5\times 10^4volt[/tex]

There is a relation between potential difference and electric field

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

[tex]E=\frac{V}{d}[/tex], here E is electric field, V is potential difference and d is distance

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

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