Question

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.

Answer 1

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 $1600\times 10^3KN/C$