How Does Distance Affect Electric Potential Near a Point Charge?

Thread starter joedirt
Start date Sep 23, 2010
Tags

Electric Electric potential Potential

AI Thread Summary

The electric potential (V) at a distance d from a point charge can be calculated using the formula V = kQ/d, where k is Coulomb's constant and Q is the charge. For a +1.0 µC charge, the potential at point A (3.0 m away) is approximately 3000 V, and at point B (6.0 m away), it is about 1500 V. The work done in moving a +0.2 µC charge from A to B can be determined by the difference in electric potential multiplied by the charge, resulting in 0.3 mJ. Understanding these calculations is essential for analyzing electric potential in electrostatics. The discussion emphasizes the relationship between distance and electric potential near point charges.

Sep 23, 2010

joedirt

Calculate electric potential at A, 3.0 m away from a point charge of +1.0 C

Calculate the electric potential at B, 6.0 m away from a point charge of +1.0 C

How much work would be done moving a +0.2 C charge between A and B

I'm not sure what equation/s to use?

Physics news on Phys.org

Sep 23, 2010

ehild

Science Advisor

Homework Helper

What is the potential at distance d from a point charge?

ehild

Thread 'Errors and significant figures'

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...

View full post »

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...

View full post »

Thread 'Explain this asphalt sheen'

This problem is two parts. The first is to determine what effects are being provided by each of the elements - 1) the window panes; 2) the asphalt surface. My answer to that is The second part of the problem is exactly why you get this affect. And one more spoiler:

View full post »