A point charge of -6.00 x 10-9C is 3.00m from point A and 5.00m from point B. Find the potential at point B. What is the potential difference between points A and B?
To find the potential at point B, we can use the formula for the electric potential due to a point charge:
V = k * (q / r)
where:
V is the electric potential
k is Coulomb's constant (k = 8.99 x 10^9 Nm^2/C^2)
q is the charge
r is the distance from the point charge
Substituting the known values:
q = -6.00 x 10^-9 C
r = 5.00 m
V_B = (8.99 x 10^9 Nm^2/C^2) * (-6.00 x 10^-9 C) / 5.00 m
Simplifying the expression:
V_B = -5.39 x 10^9 Nm^2/C
So, the potential at point B is -5.39 x 10^9 Nm^2/C.
To find the potential difference between points A and B, we need to find the potentials at both points.
Using the same formula, but substituting the distance r with the distance from point A:
q = -6.00 x 10^-9 C
r = 3.00 m
V_A = (8.99 x 10^9 Nm^2/C^2) * (-6.00 x 10^-9 C) / 3.00 m
Simplifying the expression:
V_A = -1.08 x 10^10 Nm^2/C
Now, we can find the potential difference between points A and B by subtracting the potential at point A from the potential at point B:
ΔV = V_B - V_A
ΔV = (-5.39 x 10^9 Nm^2/C) - (-1.08 x 10^10 Nm^2/C)
ΔV = -5.39 x 10^9 Nm^2/C + 1.08 x 10^10 Nm^2/C
ΔV = 5.44 x 10^9 Nm^2/C
So, the potential difference between points A and B is 5.44 x 10^9 Nm^2/C.