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

Andy Barnard asked Jan 1, 2022

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Andy Barnard asked Jan 1, 2022

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22 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

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