http://www.ck12.org Chapter 12. Electricity Version 2
Answer: Consider the diagram above; herers→eis the distance between the electron and the small charge, while
~Fs→eis the force the electron feels due to it. For the electron to be balanced in between the two charges, the forces of
repulsion caused by the two charges on the electron would have to be balanced. To do this, we will set the equation
for the force exerted by two charges on each other equal and solve for a distance ratio. We will denote the difference
between the charges through the subscripts “s” for the smaller charge, “e” for the electron, and “l” for the larger
charge.
kqsqe
r^2 s→e=
kqlqe
re^2 →lNow we can cancel. The charge of the electron cancels. The constantkalso cancels. We can then replace the large
and small charges with the numbers. This leaves us with the distances. We can then manipulate the equation to
produce a ratio of the distances.
− 3 μC
r^2 s→e=
− 3 μC
r^2 e→l⇒
r^2 s→e
r^2 e→l=
− 12 μC
− 12 μC⇒
rs→e
re→l=
√
1 μC
4 μC=
1
2
Given this ratio, we know that the electron is twice as far from the large charge (− 12 μC) as from the small charge
(− 12 μC). Given that the distance between the small and large charges is 3m, we can determine that the electron
must be located 2m away from the large charge and 1m away from the smaller charge.
Example 2
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