Answer:
potential energy at origin is [tex]2.57*10^{6} volt[/tex]
Explanation:
given data:
electric field E = 5*10^{6} N/C
at x = 43 cm, y = 28 cm
distance btween E and origin
[tex]\Delta r = \sqrt{43^2 +28^2}[/tex]
[tex]\Delta r = 51.313 cm[/tex]
potential energy per unit charge [tex]\Delta V = - Edr[/tex]
[tex]\Delta V = 5*10^6*51.313*10^{-2} J/C[/tex]
[tex]\Delta V = 2.57*10^{6} volt[/tex]
potential energy at origin is 2.57*10^{6} volt
