A uniform electric field of magnitude 270 V/m is directed in the positive x direction. A +13.0 µC charge moves from the origin to the point (x, y) = (20.0 cm, 50.0 cm).(adsbygoogle = window.adsbygoogle || []).push({});

(a) What is the change in the potential energy of the charge field system?

____ J

(b) Through what potential difference does the charge move?

____ V

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ok so

270 V/m ---->

13 x 10^-6 C

origin to (0.2m, 0.5m)

i dunno if any of this is right, just my attempt

distance from origin to point is sqrt(0.2^2 + 0.5^2)

= 0.5385 m

since the field is in the positive x direction, does this mean that the potential is only due to the y direction (since field wants charge to go right n e ways)?

change in the potential energy is the work done between so

= - q ∆V

= - (13*10^-6) (270*0.5 - 270*0)

= - (13*10^-6) (135)

then the potential difference would be the potential energy divided by the charge, so just 135 V

any help is appreciated

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# Homework Help: Potential Energy of Charge

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