How much work does the electric field do in moving a proton from a point with a potential of +125 v to a point where it is -55 v
1 answer:
The work W done by the electric field in moving the proton is equal to the difference in electric potential energy of the proton between its initial location and its final location, therefore:
where q is the charge of the proton,
, with 
being the elementary charge, and 
and 
are the initial and final voltage.
Substituting, we get (in electronvolts):
and in Joule:
where q is the charge of the proton,
Substituting, we get (in electronvolts):
and in Joule:
You might be interested in
Refer to the diagram shown below.
The component of the applied force perpendicular to the door is
F * sin(60°) = 0.866F N
Because the moment arm is 0.40 m, the torque is
(0.866F N)*(0.4 m) = 0.3464F N-m
This torque is equal to 1.4 N-m, therefore
0.3464F = 1.4
F = 4.04 N
Answer: 4.04 N
The component of the applied force perpendicular to the door is
F * sin(60°) = 0.866F N
Because the moment arm is 0.40 m, the torque is
(0.866F N)*(0.4 m) = 0.3464F N-m
This torque is equal to 1.4 N-m, therefore
0.3464F = 1.4
F = 4.04 N
Answer: 4.04 N