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babunello [35]
2 years ago
7

What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential di

fference (voltage) between them of 3.22 x 104 V
Physics
1 answer:
raketka [301]2 years ago
8 0

Answer:

E=3.22*10^6 N/C

Explanation:

From the question we are told that:

Separation Distance d=1.0cm =0.01m

Potential difference V=3.22 * 10^4 V

Generally the equation for Electric Field strength is mathematically given by

 E=\frac{v}{d}

 E=\frac{3.22*10^4}{0.01}

 E=3.22*10^6 N/C

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