The magnitude of the electric field between two parallel charged plates is 200 n/c. an electron moves to the negative plate 5.0 cm away. find the electric potential difference and the work. recall the charge of an electron is 1.602 × 10^–19 c. δv= v round work to one decimal. w= x 10^-18 j

The electric potential difference (which is also the voltage) is calculated as the magnitude of the electric field multiplied by the distance moved. For this problem, we have (200 N/C) for the electric field multiplied by 5.0 cm. We first convert 5.0 cm to 0.05 m, then multiply by 200 N/C to get 10 N-m/C = 10 J/C = 10 V.

The work itself is obtained by multiplying the potential difference by the charge of an electron (which is 1.602 x 10^-19 C). So multiplying 10 V by 1.602 x 10^-19 C, we get 1.602 x 10^-18 V-C, which is 1.602 x 10^-18 J. Rounding off to one decimal place, this is 1.6 x 10^-18 J.

The answer is b hope this