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The difference in potential between the accelerating plates of a TV set is about 24 kV. The distance between these plates is 1.5 cm. Find the magnitude of the uniform electric field in this region. Answer in units of N/C.

Respuesta :

Answer:

So electric field between the plates will be equal to [tex]1600\times 10^3KN/C[/tex]

Explanation:

We have given potential difference between accelerating plates = 24 KV = 24000 volt

Distance between the plates d = 1.5 cm = 0.015 m

We know that potential difference is given by V = Ed, here E is electric field and d is distance between plates

So [tex]24000=E\times 0.015[/tex]

E = 1600000 N/C = [tex]1600\times 10^3KN/C[/tex]

So electric field between plates will be equal to [tex]1600\times 10^3KN/C[/tex]

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