Find the coordinate where net electric field is 0

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SUMMARY

The discussion centers on determining the coordinate along the x-axis where the net electric field produced by two fixed charges, q1 = 2.1x10^-7 C at x = 20 cm and q2 = -4.00q1 at x = 70 cm, equals zero. The participant correctly identifies that the electric fields from both charges must be equal in magnitude but opposite in direction, leading to the equations E1 + E2 = 0 and E1 = -E2. The solutions found include x = -30 cm and x = 36.6666 cm, with the participant questioning the validity of multiple solutions based on the distances from the charges.

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Homework Statement



Two particles are fixed to an x axis, particle 1 of charge q1 = 2.1x10^-7 C at x = 20cm and particle 2 of charge q2 = -4.00q1 at x = 70cm. At what coordinates on the axis is the net electric field produced equal to 0?

Homework Equations





The Attempt at a Solution



First, I noted that E1 + E2 = 0, so E1 = -E2.

Then, factoring out k = 8.99x10^9, I am left with:

[itex]\frac{q_{1}}{r^{2}_{1}} = -\frac{q_{2}}{r^{2}_{2}}[/itex]
[itex]\frac{q_{1}}{r^{2}_{1}} = \frac{4q_{1}}{r^{2}_{2}}[/itex]
[itex]\frac{r_{2}}{r_{1}} = 2[/itex]

I interpreted this to mean that the charge must be twice as far from q2 as it is from q1.

So then,

x - 70 = 2(x-20)
x = -30cm


Which is what the book says, but in my method there are other solutions too, like

x - 70 = 2(20-x)
x = 36.6666

Are these other solutions valid? Can the coordinate be ANY such that it is twice as far from q2 as it is from q1?

I may have solved this in a weird way, its because I'm doing it on my own and didn't take the class yet.

Thanks again!
 
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1MileCrash said:

Homework Statement



Two particles are fixed to an x axis, particle 1 of charge q1 = 2.1x10^-7 C at x = 20cm and particle 2 of charge q2 = -4.00q1 at x = 70cm. At what coordinates on the axis is the net electric field produced equal to 0?

Homework Equations



The Attempt at a Solution



First, I noted that E1 + E2 = 0, so E1 = -E2.

Then, factoring out k = 8.99x10^9, I am left with:

[itex]\frac{q_{1}}{r^{2}_{1}} = -\frac{q_{2}}{r^{2}_{2}}[/itex]
[itex]\frac{q_{1}}{r^{2}_{1}} = \frac{4q_{1}}{r^{2}_{2}}[/itex]
[itex]\frac{r_{2}}{r_{1}} = 2[/itex]

I interpreted this to mean that the charge must be twice as far from q2 as it is from q1.

So then,

x - 70 = 2(x-20)
x = -30cm

Which is what the book says, but in my method there are other solutions too, like

x - 70 = 2(20-x)
x = 36.6666

Are these other solutions valid? Can the coordinate be ANY such that it is twice as far from q2 as it is from q1?

I may have solved this in a weird way, its because I'm doing it on my own and didn't take the class yet.

Thanks again!
Technically, you're just finding the locations at which the magnitude of the electric force due to each of the charges is equal.

If the charges have the same sign, then the fields cancel somewhere between the charges.

If the charges have opposite sign, then the fields cancel somewhere on either side of the charges.
 

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