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defon
3 years ago
10

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.
Physics
1 answer:
ryzh [129]3 years ago
7 0

Answer:

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

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 24000=E\times 0.015

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

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

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