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kodGreya [7K]
3 years ago
9

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?

Physics
1 answer:
Fudgin [204]3 years ago
3 0

Answer:

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

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 = \sqrt{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

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