Find Change in the Electric Potential

[fal01]

Homework Statement

A proton is released from rest in a uniform
electric field of magnitude 40000V/m di-
rected along the positive x axis. The proton
undergoes a displacement of 0.3 m in the di-
rection of the electric field

The mass of a proton is 1.672623×10−27 kg

A.Find the change in the electric potential
if the proton moves from the point A to B.
Answer in units of V.

B.Find the change in potential energy of the
proton for this displacement. Answer in units
of J.


C.Apply the principle of energy conservation to
find the speed of the proton after it has moved
0.5 m, starting from rest. Answer in units of
m/s.

Homework Equations

V=Ed
U=qV
Kf-Ki+u=0
k=.5mv^2
Q

The Attempt at a Solution

A.) E=40000 V/m
d=.3m
mp=1.672623*10^-27kg
q=1.60218*10^-19

V=Ed
V=(40000)(.3)=12000V ---but that is wrong

B.)
U=qV

C.)
Kf-Ki+U=0
.5mv^2+U=0

 

[PeterO]

Homework Statement

A proton is released from rest in a uniform
electric field of magnitude 40000V/m di-
rected along the positive x axis. The proton
undergoes a displacement of 0.3 m in the di-
rection of the electric field

The mass of a proton is 1.672623×10−27 kg

A.Find the change in the electric potential
if the proton moves from the point A to B.
Answer in units of V.

B.Find the change in potential energy of the
proton for this displacement. Answer in units
of J.C.Apply the principle of energy conservation to
find the speed of the proton after it has moved
0.5 m, starting from rest. Answer in units of
m/s.

Homework Equations

V=Ed
U=qV
Kf-Ki+u=0
k=.5mv^2
Q

The Attempt at a Solution

A.) E=40000 V/m
d=.3m
mp=1.672623*10^-27kg
q=1.60218*10^-19

V=Ed
V=(40000)(.3)=12000V ---but that is wrong

B.)
U=qV

C.)
Kf-Ki+U=0
.5mv^2+U=0

(a) Electric potential is measured in volts. You were given V/m and metres. Just combine them to get volts [Joules per Coulomb; which you will need for part (b)]

 

[fal01]

so what I started doing was right... (40000V/m)(.3m) =12000 V

when I put in that answer it says I am wrong :(

 

[Cade]

Potential difference is given in volts. Electric fields are measured in V/m.

E = dV/dx, so your change in potential dV = E*dx, which does match your working. However, since the electric field is doing work on the particle by pushing it, the particle is losing potential energy, not gaining it: your sign should be negative, I think.

 

[fal01]

Thanks! I was just missing the negative sign.