Work done by an electric FIELD = Work done by electric FORCE ?

AI Thread Summary
The discussion clarifies the relationship between work done by an electric field and work done by electric force, particularly for negative charges. When an electric field does negative work on a negative charge, it indicates that the electric force is opposite to the particle's displacement, leading to an increase in potential energy. Additionally, negative charges are accelerated toward points of higher electric potential, despite moving in the direction of the electric field lines. The work done by the electric field is defined as the negative change in potential energy, reinforcing the concept that potential energy increases for negative charges moving against the field. Understanding these principles is crucial for solving related physics problems effectively.
I_am_a_person
Messages
2
Reaction score
0
"Work done by an electric FIELD" = "Work done by electric FORCE"??

Homework Statement


Problem 1
If the electric field does negative work on a negative charge as the charge undergoes a displacement from position A to position B within an electric field, then the electrical potential energy:
A) is negative
B) is positive
C) increases
D) decreases
E) Cannot be determined

Homework Equations



Attempt at Solution
I assumed that "electric field does negative work" meant that the particle is moving opposite to electric field lines (if this assumption is false, and "electric field does negative work" should be interpreted as "electric FORCE does negative work", then I do not need further clarification). If so, since negative charges naturally move opposite to electric fields lines, then wouldn't potential energy decrease (D)? (The correct answer is "C": "increase")

Homework Statement


Problem 2
Negative charges are accelerated by electric fields toward points
A) At lower electric potential
B) At higher electric potential
C) where the electric field is zero
D) where the electric field is weaker
E) where the electric field is stronger

Homework Equations


ΔV = ΔU/q

Attempt at Solution
Again, I assumed that the direction of acceleration by an electric FIELD is the same direction as an electric FIELD LINE (again, if this assumption is wrong and "acceleration by an electric field" should be interpreted as "acceleration by an electric FORCE", then no further clarification is needed). If so, then if a negative charge moves in the direction of the field line, its potential energy increases (ΔU>0). Since ΔV = ΔU/q, and since ΔU>0, q<0, shouldn't ΔV be negative (A)? (The correct answer is "B": ΔV is positive)

General question: If a problem states, "an electric field does negative work on a negative charge", does the term "negative work" mean that the electric FORCE is opposite to the direction of motion or that the particle's motion is opposite the electric FIELD lines? Similarly, if a negative charge is being "accelerated electric field toward a point", is it being accelerated in the direction of the electric FIELD lines or the electric FORCE? (NOTE: The questions above were taken from a Princeton Review Subject SAT book)

Thanks in advance. And since I'm a newbie to physicsforums, I accidentally posted this outside of this HW help section, so if anyone knows how to delete posts, please enlighten me.
 
Physics news on Phys.org


Hi, person :smile:, welcome to the Forums.

You are right the electric force does work, but we say "work of the electric field".
If the field do negative work on a negative charge it means that the electric force is opposite to the displacement of the particle. The electric force F=qE is opposite to the electric field in case of negative charge. The opposite displacement has the same direction as the field, so the potential decreases, but the potential energy of the negative particle increases.

In general, the work done by a constant conservative force F when an object moves from A to B (displacement ΔrAB) is equal to the negative potential energy difference between points A and B: WAB=FΔrAB =U(A)-U(B)

ehild
 


Thanks, ehild, that really cleared things up!
 


I_am_a_person said:
Thanks, ehild, that really cleared things up!

You are welcome :smile:

ehild
 


This is the definition of the work done by an electric field (irrespective of -ve charges or +ve charges)

W_{field}=\frac{-kQq}{x}

And the work done by an electric field is always equal to :

W_{field}= -ΔP

Combining these two equations with the fact that the convention defining electric field vector,E, shows you the direction of the Force acting on a unit +ve test charge will surely help u solve the problem .
 


hms.tech said:
This is the definition of the work done by an electric field (irrespective of -ve charges or +ve charges)

W_{field}=\frac{-kQq}{x}
It is wrong anyway. But what is x?

hms.tech said:
And the work done by an electric field is always equal to :

W_{field}= -ΔP

The work of the electric field on a charged body depends on the charge. What do you denote by P?

ehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top