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.
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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
 
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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
 
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