10. The following three point charges are placed on the x axis: +5.0 mC at x=0; +4.0 mC at x= -40 cm; and +6.0 mC at x= +80 cm. Find (a) the electric field magnitude and direction and (b) the electric potential at x= +10 cm.
a) E = k(5E-6)/.102 +k(4E-6)/.502 - k(6E-6)/.702 = +4.5E6 N/C (in +x direction)
I was just wondering, how did the teacher get .102, .502 and .702 as the radius's.
x is 0.70 m from the 6 mC charge, 0.50 m from the 4 mC charge, and 0.10 m from the 5 mC charge
I also do not understand why the number "0.002" was added to those distances (in meters). It appears to be a misprint.
To calculate the electric field magnitude at a specific point, we need to consider the distance between the point charge and that point. In this case, the point charges are placed on the x-axis.
For the first point charge at x = 0, the distance (radius) between the charge and the point in question (x = +10 cm) is calculated as follows:
radius = +10 cm - 0 = +10 cm = 0.1 m
For the second point charge at x = -40 cm, the distance (radius) between the charge and the point in question (x = +10 cm) is calculated as follows:
radius = +10 cm - (-40 cm) = +50 cm = 0.5 m
For the third point charge at x = +80 cm, the distance (radius) between the charge and the point in question (x = +10 cm) is calculated as follows:
radius = +10 cm - (+80 cm) = -70 cm = 0.7 m
So, the teacher used 0.1 m, 0.5 m, and 0.7 m as the radii for calculating the electric field magnitude at x = +10 cm.