Electric Field Between two Charges Equals 0

[Hypnos_16]

Homework Statement

A 3.17 μC and a -2.19 μC charge are placed 3.65 cm apart. At what point along the line joining them is the electric field zero? Assume that the first charge is at the origin and the second charge is at +3.65 cm.

Homework Equations

E = K(q) / r²
Ep = E1 + E2
0 = E1 + E2
-E1 = E2
Since one is negative and the other is positive they would really work together making it E1 + E2, then -E1 = E2

The Attempt at a Solution

Ep = E1 + E2
0 = E1 + E2
-E1 = E2
-k(q1) / x² = k(q2) / (r - x)²
-(q1) / x² = (q2) / (r² - 2rx + x²)
-(q1)(r² - 2rx + x²) = (q2)(x²)
-(3.17e-6)(0.0365² - 2(0.0365)x + x²) = (-2.19e-6)(x²)
-(3.17e-6)(1.33e-3) - (7.30e-2)x + (x²) = (-2.19e-6)(x²)
-(3.17e-6)²2 + (7.30e-2)x - (4.22e-9) + (2.19e-6)x² = 0
(-9.80e-7)²2 + (7.30e-2)x - (4.22e-9) = 0
x = -b +/- √[b² - 4ac / 2a
x = [-(7.30e-2) +/- √(7.30e-2)² - (4)(-5.36e-6)(-4.22e-9)] / 2(-9.80e-7)
x = [-(7.30e-2) +/- (5.33e-3)] / [-1.96e-6]

Sorry if that's hard to read. But after i did all this, i got answers way to big to even be close. They were easily in the 10s of thousands. What did i do wrong?

 

[SammyS]

You mentioned that (for a point between the charges) the E-field for both charges is in the same direction. Actually, it points left.

A line is infinitely long, so find a place along the x-axis, that's not between the charges.

Do you suppose that point is to the right, or is it to the left of the two charges?

 

[Hypnos_16]

After i got an answer the two points where to the right of the two charges.

 

[SammyS]

What two points?

There is one point at which the E-field is zero.

That point is to the right ot the two charges. Why?

 

[lightgrav]

you should only have one place that their field contributions cancel.
(the other "zero" is at infinity)
but yes, it's farther from the big charge sqrt(3.19/2.17) times as far.

 

[Hypnos_16]

What two points?

There is one point at which the E-field is zero.

That point is to the right ot the two charges. Why?

I mean, after doing Quadratic i ended up with two values, both where larger than 3.65 cm. That's what i mean when i said i have two points.

It's to the right because we the field is going to the right. We were taught that you always imagine what a proton would do should it enter the field, and in this case it would be repelled from the positive charge and attracted to the negative charge to the right of it. Thus the point where the field is zero is the right of the two charges.

 

[lightgrav]

It depends on which place has more charge.

 

[Hypnos_16]

Alright, so we know that it's to the right, I don't see how this is helping solving the problem i have.

 

[lightgrav]

solve for q2/q3 = (x-d)^2 / (x)^2 ... then take the root of both sides.

 

[Hypnos_16]

solve for q1/q2 = (x-d)² / (x)² ... then take the root of both sides.

q1 / q2 = (x - d)² / (x)²
3.17 / -2.19 = (x - d)² / (x)²
√[3.17 / -2.19 = (x - d)² / (x)²]
1.20 = (x - d) / x
1.20x = x - d
1.20(0.0365) = (0.0365 - d)
0.0438 = 0.0365 - d
-0.0438 = -0.0365 + d
d = -0.0073 m => -0.73 cm Something like that then?

 

[lightgrav]

the big charge has to be farther away, so its influence is spread out more.
I thought "x" was the variable , and "d" was the constant.
(I would've done 0.4617 = 1 - d/x)

 

[Hypnos_16]

Where did you get the 0.4617 From? Everything else i can see.

 

[lightgrav]

E3 = 3/(x)^2 . . . E2 = 2/(x-d)^2
=> sqrt(2/3) = (x-d)/x = 1 - d/x
oops. I needed parentheses in my calculator!

 

[Hypnos_16]

I'm assuming that the 3 and the 2 that you're using are the charges? Then why not add the Negative to the two? or is it irrelevant in this case

E3 = 3/(x)^2
E2 = 2/(x-d)^2

3.17 / x² = 2.19 / (x - d)²
√(2.19 / 3.17) = (x - d) / x
√(2.19 / 3.17) = (1) - (d / x)
0.8311 = 1 - (d / x)
0.8311 - 1 = -d / x
-0.1688 = -d / x
-0.1688x = -d
-0.1688x = -(0.0365)
x = -0.0365 / -0.1688
x = 0.216m => 21.6 cm