How Is Work Calculated When Moving a Charge in an Electric Field?

AI Thread Summary
The discussion focuses on calculating the work done in moving a 0.2 µC charge within an electric field created by a 10 µC charge at the center of a square. The key equations involved are work or change in potential energy (ΔU=q*V) and electric potential (V=kq/r). The challenge arises from the lack of a specific distance (r) between the charges, leading to assumptions about the square's dimensions. It is emphasized that the work done is dependent on the difference in potential between the starting and ending points, suggesting the use of ΔU=q*ΔV for a more accurate calculation. The conversation highlights the importance of understanding electric forces as conservative, impacting how work is computed in this context.
smartdesk
Messages
3
Reaction score
0

Homework Statement


What is the work done in moving a 0.2 µC point charge from corner a to b of a square abcd, when a 10 µC charge exists at the center of the square?


Homework Equations


Work or ΔU=q*V
V=kq/r


The Attempt at a Solution


V=kq/r
V=(9.0x10^9)(0.2x10^-6)/r

This is where I get stuck because there is no r given. We can assume that all the sides of the square are equal and the distance to the 10 µC charge is half of that. But I feel like an actual value should be given.
 
Physics news on Phys.org
Assume some length of the square side, see if it doesn't cancel out in the final result.
 
I assumed the length of he side of the square to be 0.3m, but it doesn't cancel out. So if the side were 3m, electric potential and work done would be:
V=kq/r=(9.0*10^9)(0.2*10^-6)/3=600 N

ΔU=q*V=(0.2*10^-6)(600)=1.2*10^-4 J
 
i meant 3m not 0.3 m
 
Hi smartdesk

Work out the potential energy of the system when the charge is at a and then again when the charge is at b, and then compare them.
Just use r for the distance.

Notice anything?
 
smartdesk said:
Work or ΔU=q*V

That's not correct. Electric force is conservative, which means work done depends on the difference between potential in the starting and ending point. So the more correct way of expressing it would be ΔU=q*ΔV. Now think what ΔV is.

Edit: ap123 answered while I was composing the answer, but we are aiming at exactly the same thing.
 
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