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What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 3.22 x 104 V
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What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 3.22 x 104 V

User David Vrba
by
4.4k points

1 Answer

3 votes

Answer:


E=3.22*10^6 N/C

Step-by-step 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=(v)/(d)


E=(3.22*10^4)/(0.01)


E=3.22*10^6 N/C

User Zimi
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4.8k points
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