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olya-2409 [2.1K]
3 years ago
6

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
Physics
1 answer:
sleet_krkn [62]3 years ago
5 0

Answer:

d=2.7m

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=\frac{V}{E}

d=\frac{23.0*10^3}{8.53 * 10^3 V/m}

d=2.7m

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<em><u>Incomple</u></em><em><u>te</u></em><em><u> </u></em><em><u>ques</u></em><em><u>tion</u></em><em><u>!</u></em>

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