How does the Electric Potential change when -q goes to -2q?

AI Thread Summary
The discussion centers on understanding how electric potential changes when a negative charge of -q is replaced by -2q. A positive charge +q is located at x=+a, while the negative charge is at x=-a, and the goal is to find the potential along the y-axis. Electric potential is a scalar quantity, meaning it has no direction, and the total potential from multiple charges is simply the sum of the potentials from each charge. The original poster notes that if the charge were -q instead of -2q, the potential would be zero, indicating a need for clarity on the impact of the charge's magnitude. The conversation emphasizes the straightforward nature of calculating electric potential from multiple sources.
durhamisnuts
Messages
1
Reaction score
0

Homework Statement



Imagine that a particle with positive charge +q is placed on the x-axis at x=+a and a particle with negative charge -2q is placed on the x-axis at x= -a. Find the potential at all points along the y axis.



Homework Equations



delta phi = -E dot dr

The Attempt at a Solution



This isn't a homework problem, it's just a problem in the book I'm working on to get a better understanding of electric potential. I was wondering if someone could explain electric potential to me, and i figured the easiest way would be with a problem. The answer is given in terms of k, q, a, and y.

I know that had the it been -q instead of -2q the answer would be zero, but I am not sure how the physics works when it is -2.
 
Physics news on Phys.org
welcome to pf!

hi durhamisnuts! welcome to pf! :wink:

potential is a scalar

(unlike electric field, which is a vector)

ie, it's just a number, it has no direction …

so to get the potential from two or more sources, you find the potential from each, and add them

it's that easy! :smile:
 
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