17. Potential due to a point charge: What is the electric potential 17.0 cm from a 3.50 μC point charge?
V = -k Q/R
k is the Coulomb's Law constant
= 8.99*10^9 N*m^2/C^2
Q = 3.50*10^-6 C
R = 0.17 m
Solve for V, which will be in J/C units
So because of the negative sign in the question would the answer end up being negative?
I meant to say the negative sign in the equation.
And Thank you very much.
I was wrong about the negative sign. Relative to zero voltage at infinity, the potential at nonzero distances will be positive.
http://en.wikipedia.org/wiki/Electric_potential
My brain isnt awake yet
To find the electric potential due to a point charge, you can use the formula:
V = k * q / r
Where:
V is the electric potential (in volts),
k is the electrostatic constant (9 x 10^9 N·m^2/C^2),
q is the charge of the point charge (in coulombs), and
r is the distance from the point charge (in meters).
In this case, the point charge is 3.50 μC (microcoulombs), which can be converted to coulombs by dividing by 10^6:
q = 3.50 μC / 10^6 = 3.50 × 10^-6 C
The distance from the point charge is 17.0 cm, which needs to be converted to meters by dividing by 100:
r = 17.0 cm / 100 = 0.17 m
Now, you can plug these values into the formula to calculate the electric potential:
V = (9 × 10^9 N·m^2/C^2) * (3.50 × 10^-6 C) / (0.17 m)
By performing the calculations, you can find the electric potential at a distance of 17.0 cm from the point charge.