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How far apart are two conducting plates that have an electric field strength of 8.53 x 103 V/m between them, if their potential difference is 23.0 kV
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How far apart are two conducting plates that have an electric field strength of 8.53 x 103 V/m between them, if their potential difference is 23.0 kV

User Dyoo
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3.6k points

1 Answer

3 votes

Answer:


d=2.7m

Step-by-step explanation:

From the question we are told that:

Electric Field strength
E=8.53 * 10^3 V/m

Potential difference is
V= 23.0 kV

Generally the equation for distance is mathematically given by


d=(V)/(E)


d=(23.0*10^3)/(8.53 * 10^3 V/m)


d=2.7m

User Psychobunny
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3.4k points
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