Work and Force on a positively charged particle in a capcitor

[UIUCstudent]

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So let's say we have a
http://www.screencast.com/users/trinhn812/folders/Jing/media/547f1ab2-1520-439e-a326-c408d3d799a2

Electric field between the two parallel charged plates is constant and uniform, so is the force at point A, B, and C the same (assuming the charge of both plates are the same in magnitude)?

Also when a charged particle is traveling along an equipotential line, there is no force and work acting on it? Why is that?

 

[flodhi1]

So let's say we have a
http://www.screencast.com/users/trinhn812/folders/Jing/media/547f1ab2-1520-439e-a326-c408d3d799a2

Electric field between the two parallel charged plates is constant and uniform, so is the force at point A, B, and C the same (assuming the charge of both plates are the same in magnitude)?

Also when a charged particle is traveling along an equipotential line, there is no force and work acting on it? Why is that?

So the Electric field is kq1/r^2 and Force is kq1q2/r^2. Thus Force is simply (F=qE) Where F is force, q is charge and Electric field is E. Thus based on that if the ELECTRIC field is constant and the charges are all the same than the Force SHOULD be the same.
Concerning your other question Work can be defined as Potential energy. It's kind of like P.E = mgh which determines the amount of work required to go to a certain height. Similarly when moving in equipotential lines you're not changing the "h" and by the way P.E is = kq1q2/r and V= kq1/r thus Work = q delta V. So if you're on the same equipotential line think of it as you're not changing "h" in the mgh. Thus if you're not changing the potential energy the Work is 0.

 

[UIUCstudent]

So the Electric field is kq1/r^2 and Force is kq1q2/r^2. Thus Force is simply (F=qE) Where F is force, q is charge and Electric field is E. Thus based on that if the ELECTRIC field is constant and the charges are all the same than the Force SHOULD be the same.
Concerning your other question Work can be defined as Potential energy. It's kind of like P.E = mgh which determines the amount of work required to go to a certain height. Similarly when moving in equipotential lines you're not changing the "h" and by the way P.E is = kq1q2/r and V= kq1/r thus Work = q delta V. So if you're on the same equipotential line think of it as you're not changing "h" in the mgh. Thus if you're not changing the potential energy the Work is 0.

Ahh I see. I got caught up in the "r" variable. But it shouldn't be taken into account unless the electric field lines aren't constant. Thanks!