The electric field between two parallel plates has a magnitude of 1250 N/C. The positive plate is 0.05 m away from the negative plate. The electric potential difference between the plates, rounded to the tenths place, is V.

ANSWERS

2016-04-17 13:05:57

Electric potential difference or the voltage is the difference of the electric potential of two points or plates. To determine the electric potential difference between the plates, we need to know the fact that the units Newtons per Coulumb (N/C) is equivalent with Volts per meter (V/m). So, we simply multiply the distance between the plates to the magnitude of the electric field of the two plates. We do as follows: Electric potential difference = Electric field strength (distance of the plates) Electric potential difference = 1250 V/m (0.05 m) Electric potential difference = = 62.5 V Therefore, the potential difference between the plates which are 0.05 m apart would be 62.5 V.

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