Electric Potential: Find V between 6.17 nC & -4.32 nC Charges

AI Thread Summary
To find the electric potential at a point midway between two charges of 6.17 nC and -4.32 nC separated by 29.1 cm, the relevant equation is V = kq/r. The acceleration due to gravity is not relevant to the calculation of electric potential in this context. The distance r for each charge should be half of 29.1 cm, which is 14.55 cm. The signs of the charges must be considered when calculating their contributions to the total potential, but the distance is treated as positive. Ultimately, the total electric potential is the sum of the potentials from each charge at the midpoint.
shaka23h
Messages
38
Reaction score
0

Homework Statement



Two pont charges of magnitude 6.17 nC and -4.32 nC are separated by 29.1 cm. The acceleration of gravity is 9.8 m/s^2. the Coulomb constant is 8.98755 x 10 ^9 N m^2/C^2. What is the electric potential at a point midway between the charges? Answer in units of V.

Homework Equations



well I know Columb's law obviously comes into effet. If its just finding the electric force of repulsion between the 2 object it would be ez. But the potential? IS the acceleration of gravity even relevant in this problem?

The Attempt at a Solution




I know that (electric potential ) V = kq/ r and Columb's equation is F = Kq1q2/r^2. How can I manipulate both of these equations?

Thanks
 
Physics news on Phys.org
I don't know why they threw in gravity, just add up the voltages. Coulombs law does not need to be manipulated
 
So do I just use V = Kq/r and solve for 6.17 and -4.32 making sure that I keep the signs? and then add these up? The question asked for the midway between charges how does this come into effect? is my r value 1/2 of 29.1 cm?

Let me know
 
Yes, but positive/negative orientation of distance doeesnt matter. The distance is positive for both voltages. Needless to say, youll wind up with a positive answer
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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