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3 years ago

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Two parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude

of the electric field strength between them, if the potential 7.35 cm from the zero volt plate (and 2.65 cm from the other) is 533 V

1 answer:

aivan3 [116]3 years ago

7 0

Answer:

$E=6.8Kv/m$

Explanation:

From the question we are told that

Distance b/w plate $d=10cm=>0.1m$

P_1 Potential at 7.35 $V=533v$

Generally the equation for electric field at a distance is mathematically given as

$E=\frac{v}{d}$

$E=\frac{533}{7.85*10^-^2}$

$E=6789.808917$

$E=6.8*10^3$

$E=6.8Kv/m$

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Answer:

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Hence, the voltage across the 1.2 square cm wire=63.8 V

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