confused: work done by electric field

[wooki]

i am sure confused about the work done by an electric field. i've read from different sources that the displacement has to oppose the direction of the electric field and then some say work is done when it goes in the same direction too... 🙁

so i made my own schematic:

- the black arrows going to the right is the electric field
- the blue arrow is the path of a positive charge (+Q)
- for each case, is there work done by E? and is there a change in electric potential (i am assuming this is redundant)

thanks!

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[novawildcat]

i am sure confused about the work done by an electric field. i've read from different sources that the displacement has to oppose the direction of the electric field and then some say work is done when it goes in the same direction too... 🙁

so i made my own schematic:

- the black arrows going to the right is the electric field
- the blue arrow is the path of a positive charge (+Q)
- for each case, is there work done by E? and is there a change in electric potential (i am assuming this is redundant)

thanks!

In order for work to be done on the particle, it must experience a force. Force and electric field are related by the equation F=Eq. Notice, there is no cross or dot product involved. The force vector can be found by simply multiplying the electric field vector by a scalar. This means a charged particle in an electric field always experiences a force. Now look at the definition of work. W=Fd. Work is a scalar, thus you need a dot product to get a scalar from two vectors. W=Fdcos(theta). You know a charged particle in an electric field experiences a force. Thus work may or may not be done depending on the cos of the angle between the force vector and the displacement vector. What is the angle between the force vector and the displacement vector in each of your pictures?


I think you are confusing this with magnetic fields and charged particles. The force exerted on a charged particle as it moves in a magnetic field is F=qvxB. Notice how this force is dependent on a cross product, i.e. the sine of the angle between the magnetic field vector and the velocity vector is important. Depending on the angle between those two vectors, there may or may not be an applied force, hence work may or may not be done.


I know I was a bit long winded, but if you have a firm grasp on simple concepts like vectors, it can go a long way to explaining a bunch of physics. Don't let the equations trick you. If you know vectors, scalars, dot products, and cross products you're golden.

 

[wooki]

hey that really helped. i think i did kind of confused with magnetism in there because work is sin and magnetism force is cosine.

so if i recalibrated my concepts right.. then

a. no work done?, no change in potential
the theta here is 90 degrees, cos 90 = 0.

b. work done, change in potential
the theta here is 0 degrees, cos 0 = 1.

c. work done, change in potential
the theta here is 180, cos 180 = -1.

now if i were to take E and change it to B...
a. work done by the magnetic field
sin 90 = 1

b. no work done by the magnetic field
sin 0 = 0

c. no work done by the magnetic field
sin 180 = 0

 

[novawildcat]

hey that really helped. i think i did kind of confused with magnetism in there because work is sin and magnetism force is cosine.

so if i recalibrated my concepts right.. then

a. no work done?, no change in potential
the theta here is 90 degrees, cos 90 = 0.

b. work done, change in potential
the theta here is 0 degrees, cos 0 = 1.

c. work done, change in potential
the theta here is 180, cos 180 = -1.

now if i were to take E and change it to B...
a. work done by the magnetic field
sin 90 = 1

b. no work done by the magnetic field
sin = 0

c. no work done by the magnetic field
sin 180 = 0

Yes it looks like you understand, although I'm not sure if you realize you have a typo in the begining. Works would come from a cosine and force from a magnetic field would be sine. Just remember, a scalar quantity from two vectors has to be a dot product and that means cosine. If you get a vector from two other vectors (assuming we aren't talking about addition or subtraction here) then there has to be a cross product invloved, and that means sine between the 2 vectors. Mcat prep books skimp out on vectors for reasons I don't know. If you understand dot products, cross products and scalars it carriers you a long way. There's no need to memorize any right hand rules etc. Understanding them makes answer questions like yours, which are mcat favorites, a cinch.


EDIT: I just realized, check you theta values again. Make sure you know which way the force vector is going. It looks like you understand the concept, but just need to recheck your work.

 

[afruff23]

hey that really helped. i think i did kind of confused with magnetism in there because work is sin and magnetism force is cosine.

so if i recalibrated my concepts right.. then

a. no work done?, no change in potential
the theta here is 90 degrees, cos 90 = 0.

b. work done, change in potential
the theta here is 0 degrees, cos 0 = 1.

c. work done, change in potential
the theta here is 180, cos 180 = -1.

now if i were to take E and change it to B...
a. work done by the magnetic field
sin 90 = 1

b. no work done by the magnetic field
sin 0 = 0

c. no work done by the magnetic field
sin 180 = 0

for an E field:

a. no work done

b. negative work done, because electric field lines go from positive to negative. This means that (-) is on the right side, so the (+) particle has a force towards the right. Since the (+) particle moves left, the work is negative.

c. positive work done, see b.

for a B field (magnetic:

a. no work, see the right hand rule (NOTE: right hand rule works for negative particles, so you must either use your left hand instead or use the opposite result of the right hand rule). Using the RHR, B field is to the right, velocity is up, so magnetic force for a NEGATIVE particle is into the paper. Since the particle is POSITIVE, the magnetic force is out of the paper. Since the force and velocity are perpendicular, then work done is 0.

b. no work, there is no component of the velocity of the (+) particle that is orthogonal (e.g. perpendicular) to the magnetic field, so there is no force at all.

c. no work, there is no component of the velocity of the (+) particle that is orthogonal (e.g. perpendicular) to the magnetic field, so there is no force at all.

 

[Rabolisk]

Magnetic field never does work on a charged particle. However, I wonder, why did you feel the need to revive a 6 month old topic?