The potential difference between the two ends of the circuit is the electric potential difference. The electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.
An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.
The given data in the problem is given by;
E is the electric field = (200 N/C)
d is the distance = 5.0 cm.=0.05 m
Q is the charge of electrons= 1.602 x 10^-19 C
The formula for electric potential is given by;
[tex]\rm V=Ed[/tex]
[tex]\rm V=Ed \\\\ \rm V=200 \times 0.05 \\\\ \rm V= 10 \frac{Nm}{C} = 10 \frac{J}{C} = 10 V.[/tex]
The work is defined as the product of the potential difference and charge of an electron.
[tex]\rm W= 10 \times 1.602 x 10^{-19} \\\\\ \rm W= 1.6 x 10^{-18 }J[/tex]
Hence the electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.
To learn more about the electric field refer to the link;
https://brainly.com/question/15071884
