http://www.ck12.org Chapter 13. Electricity Version 2
Before we solve for the electric field by plugging in the values, we convert all of the values to the same units.
4 .0mm×
1m
1000mm
=.004m− 2. 0 μC×1C
1000000 μC=− 2. 0 × 10 −^6 C
Now that we have consistent units we can solve the problem.
E=
kq
r^2=
9 × 109 Nm^2 /C^2 ×− 2. 0 × 10 −^6 C
(.004m)^2=− 1. 1 × 109 N/C
To solve for the force at the point we will use the equation
F=EqWe already know all of the values so all we have to do is convert all of the values to the same units and then plug in
the values.
− 8. 0 μC×1C
1000000 μC=− 8. 0 × 10 −^6 C
F=Eq=− 8. 0 × 10 −^6 C×− 1. 1 × 109 N/C=9000N