A point charge 1.2 nC is located at a point with coordinates (x0 = 2, y0 = 3, z0 = 3).Calculate the potential difference between two points with coordinates (in theCartesian system) (2,2,3) and (-2,3,3
Question
A point charge 1.2 nC is located at a point with coordinates (x0 = 2, y0 = 3, z0 = 3).Calculate the potential difference between two points with coordinates (in theCartesian system) (2,2,3) and (-2,3,3
Solution
To calculate the potential difference between two points, we can use the formula:
ΔV = k * (q / r1 - q / r2)
where ΔV is the potential difference, k is the Coulomb's constant (8.99 x 10^9 N m^2/C^2), q is the charge, and r1 and r2 are the distances between the charge and the two points.
First, let's calculate the distance between the charge and the point (2,2,3):
r1 = sqrt((x1 - x0)^2 + (y1 - y0)^2 + (z1 - z0)^2) = sqrt((2 - 2)^2 + (2 - 3)^2 + (3 - 3)^2) = sqrt(0 + 1 + 0) = sqrt(1) = 1
Next, let's calculate the distance between the charge and the point (-2,3,3):
r2 = sqrt((x2 - x0)^2 + (y2 - y0)^2 + (z2 - z0)^2) = sqrt((-2 - 2)^2 + (3 - 3)^2 + (3 - 3)^2) = sqrt((-4)^2 + 0 + 0) = sqrt(16) = 4
Now, we can substitute the values into the formula:
ΔV = k * (q / r1 - q / r2) = (8.99 x 10^9 N m^2/C^2) * (1.2 x 10^-9 C / 1 - 1.2 x 10^-9 C / 4) = (8.99 x 10^9 N m^2/C^2) * (1.2 x 10^-9 C / 1 - 1.2 x 10^-9 C / 4) = (8.99 x 10^9 N m^2/C^2) * (1.2 x 10^-9 C - 0.3 x 10^-9 C) = (8.99 x 10^9 N m^2/C^2) * (0.9 x 10^-9 C) = 8.091 x 10^9 N m^2/C
Therefore, the potential difference between the two points is 8.091 x 10^9 N m^2/C.
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