Potential Energy of a Charged Particle in a Uniform Electric Field

[brunie]

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).
(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 don't know 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

 

[brunie]

hmmmmmmmmmmmm

 

[brunie]

........

 

[e(ho0n3]

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)?

It seems your problem is your understanding of potential energy. How would you define the electrical potential energy at some point given the fact that you know that the electric field is uniform in one direction?

 

[KingNothing]

Wrong. Charges always want to move with the field. Potential is dependent only upon the change in x-value, NOT y.