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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

10 votes

Answer:


E=6.8Kv/m

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


E=(533)/(7.85*10^-^2)


E=6789.808917


E=6.8*10^3


E=6.8Kv/m

answered
User Sarvan Kumar
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