Two Charges - Electric Potential Question

AI Thread Summary
The discussion centers on calculating the electric potential at point C due to two fixed charges of 4.0μC each, positioned 4.0 cm apart, with V set to 0 at infinity. The initial calculation for the potential at point C was incorrectly derived, leading to confusion about the correct value. The correct distance should be converted to meters, as 4 cm equals 4×10^-2 m, which affects the calculations. Additionally, there is uncertainty regarding how to compute the work done when bringing a third charge from infinity, as the distance to point C needs clarification. The thread emphasizes the importance of accurate unit conversion in electric potential calculations.
pierretong
Messages
3
Reaction score
0

Homework Statement



Two charges q = 4.0μ C are fixed in space a distance d = 4.0 cm apart, as shown in the figure. With V = 0 at infinity, what is the electric potential at point C?

You bring a third charge q = 4.0μC from infinity to C. How much work must you do?

What is the potential energy U of the three-charge configuration when the third charge is in place?

Diagram shows (let me use x-y coordinate terms to describe this). Charge q at the origin. Another charge q at (4 cm, 0 cm) and point C at (2 cm, 2 cm)


Homework Equations



Electric Potential = kQ/r

The Attempt at a Solution



That's the right equation for the first part correct? I get 1.27*10^4 for the electric potential for one of the charges on point C. When you double that, you get 2.5 * 10^4 V and that should be the answer? (online homework system says that it is wrong).
What am I doing wrong here?

Also I don't understand the second part. If you are moving a charge from infinity, how can you calculate the work? You don't know the distance?
 
Physics news on Phys.org
pierretong said:

Homework Statement



Two charges q = 4.0μ C are fixed in space a distance d = 4.0 cm apart, as shown in the figure. With V = 0 at infinity, what is the electric potential at point C?

You bring a third charge q = 4.0μC from infinity to C. How much work must you do?

What is the potential energy U of the three-charge configuration when the third charge is in place?

Diagram shows (let me use x-y coordinate terms to describe this). Charge q at the origin. Another charge q at (4 cm, 0 cm) and point C at (2 cm, 2 cm)


Homework Equations



Electric Potential = kQ/r

The Attempt at a Solution



That's the right equation for the first part correct? I get 1.27*10^4 for the electric potential for one of the charges on point C. When you double that, you get 2.5 * 10^4 V and that should be the answer? (online homework system says that it is wrong).
What am I doing wrong here?

Also I don't understand the second part. If you are moving a charge from infinity, how can you calculate the work? You don't know the distance?
Hello pierretong. Welcome to PF !

I see that this is your 2nd post & that you posted your first thread yesterday?

I hope DocAl's explanation was helpful, and that you were you able to solve your problem.

As for this thread:
4 cm = 4× 10-2 m, not 4× 10+2 m.​
Your answer is in error by some power of 10.
 
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