Calculating external work done in electric field

In summary, a region of space has a uniform electric field of 10.0 V/m due east. A small metal sphere with charge of +50.0 nC is moved at a constant speed by an external force 10 cm north, 50 cm east, 20 cm south, and 50 cm east. The total work done by the applied force is the sum of the individual work required for each movement, and only the east movements need to be considered as the force perpendicular to the field does no work. The change in potential for the sphere is equal to the work done, which is negative if the sphere is moving in the direction of the force.
  • #1
winterwind
29
0

Homework Statement


A region of space has a uniform electric field of strength 10.0 V/m due east. A small metal sphere with charge of +50.0 nC is in the field. There sphere is moved at a constant speed by an external force 10 cm north, 50 cm east, 20 cm south, and 50 cm east. How much work was done in total by the applied force?


Homework Equations



W = qEs

The Attempt at a Solution



W = qE, so I tried to plug in the charges and electric field information. But I don’t know how to apply the fact that the electric field is directed due east. I can calculate the work if it were moved AGAINST the electric field, but I don’t know how to calculate it if it were moved in the same direction as the electric field or perpendicular to it, as in this case.

The total work is just the sum of all the individual work required to move the charge each time.

Thanks for your help!
 
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  • #2
If you have a force that's perpendicular to the direction of motion it does no work. It's the same for this. W=qEs, but F=qE, so W=Fs. If you apply the force perpendicular to the field, it does no work. So you only have to consider the east movements.
 
  • #3
Thanks.

The work is the same if I move it x amount east or x amount west, with the field due east?

Part two of the question: What is its change in potential?

Attempt:

Would the change in potential just be the amount of work done? Or is it zero since it is in the same electric field?
 
  • #4
Yes, the magnitude of work will at least be the same. I'm not 100% sure about it being positive or negative, but it's early in the morning and I can't think properly.

No, Work= negative change in potential energy. Potential energy is related to potential.
 
Last edited:
  • #5
If you are moving in the direction of the force then you have (-) work.

Like for instance gravity going down gives back -work as kinetic energy.

With work being given by the relationship:

W = q*ΔV

and your field given as 10v/m, if you move in the direction of the field you are moving toward lower voltage at the rate of 10v/m. So if you move 50cm + 50 cm toward negative voltage, your voltage will be lower by what amount? Since the ΔV is negative then your work should be negative shouldn't it?
 

Related to Calculating external work done in electric field

1. What is external work done in an electric field?

The external work done in an electric field refers to the amount of energy transferred to or from a charged particle as it moves through the field. It takes into account the force applied to the particle and the distance it travels in the field.

2. How is external work done calculated in an electric field?

The external work done in an electric field is calculated by multiplying the magnitude of the electric field by the component of the particle's displacement in the direction of the field. This can be represented by the equation W = F * d * cosθ, where W is the external work done, F is the force applied to the particle, d is the displacement, and θ is the angle between the force and displacement vectors.

3. What is the unit of measurement for external work done in an electric field?

The unit of measurement for external work done in an electric field is joules (J). This is the standard unit for measuring energy.

4. Can external work done in an electric field be both positive and negative?

Yes, external work done in an electric field can be both positive and negative. If the force and displacement vectors are in the same direction, the work done will be positive. If they are in opposite directions, the work done will be negative.

5. How does external work done in an electric field affect the movement of charged particles?

The external work done in an electric field can either increase or decrease the kinetic energy of charged particles, depending on the direction of the force and displacement vectors. This can cause the particles to speed up, slow down, or change direction as they move through the field.

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