Question

What is the potential energy at the origin due to an electric field of 5 x 10^6 N/C located at x=43cm,y=28cm?

Answer 1

potential energy at origin is
`2.57*10^(6) volt`

Step-by-step explanation:

given data:

electric field E = 5*10^{6} N/C

at x = 43 cm, y = 28 cm

distance btween E and origin

`\Delta r = √(43^2 +28^2)`

`\Delta r = 51.313 cm`

potential energy per unit charge
`\Delta V = - Edr`

`\Delta V = 5*10^6*51.313*10^(-2) J/C`

`\Delta V  =  2.57*10^(6) volt`

potential energy at origin is 2.57*10^{6} volt