Work done by E field to move electron from negative plate to postive.

[whynot314]

The facing surfaces of two large parallel conducting plates separated
by 10.0 cm have uniform surface charge densities that are equal in magnitude but
opposite in sign. The difference in potential between the plates is 500 V. An electron is released from rest next to the negatively charged surface. Find the work done by the electric field on the electron as the electron moves from the release point to the positive plate.

Ok so I am getting -8E-17 J

by ΔU=qΔV, W=qΔV

and I know ΔV is positive because we are going form low potential to high potential therefore Δv=500V

so i am getting W=(-1.6E-19)(500)=-8E-17 however I am concerned because I am seeing a positive work somewhere else. and that just does not make sense to me. or can someone explain to me why this would be positive work?

 

[kevinferreira]

The electron is released next to the negative plate. The electron will be attracted to this plate, so to get it to the other plate you must give it energy, and therefore work is positive.

 

[whynot314]

I thought the electron would be repelled by the negative plate since they have the same charge??

 

[whynot314]

Ok I believe I have everything sorted out now, I would be doing negative work If an EXTERNAL FORCE was applied to move the electron to the positive plate, the problem asks for the work done the the ELECTRIC FIELD, and since the electric field is going to the opposite direction the of the motion of the electron w=qvcos(180)= -+-=+work.

 

[Nickie]

I would like to clarify that the work done by the electric field in this scenario is indeed positive. This may seem counterintuitive because the electron is moving from a negatively charged plate to a positively charged plate, but it is important to remember that work is defined as the force applied over a distance. In this case, the electric field is providing a force on the electron, pushing it towards the positively charged plate.

When the electron is released from rest, it has no kinetic energy, so all of the work done by the electric field goes towards increasing the electron's potential energy. This potential energy is directly related to the voltage difference between the plates, as shown in the equation W=qΔV. Since the electron is moving towards a higher potential, the work done by the electric field is positive.

In summary, the work done by the electric field is positive because it is providing the force that moves the electron from the negative plate to the positive plate, increasing its potential energy. I hope this explanation helps clarify any confusion.