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 electric field strength (in kV/m) between them, if the potential 9.00 cm from the zero volt plate (and 1.00 cm from the other) is 440 V

4888.89V/m

Explanation:

Given the following :

Distance from zero volt plate = 9cm

Electrical potential at the distance is given as 440 V

The electric field strength measures the numerical intensity of an electric field at a particular point. It ua related by the ratio of the potential difference between points of speration in of plates.

Electric field strength = potential difference / distance

Electric field strength = (440V / (9/100)m)

Electric field strength = 440V / 0.09m

Electric field strength = 4888.89V/m

the answer would be c

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