Electric Field Zero: Where is the Point in Space?

[exitwound]

Homework Statement

Homework Equations

E=kQ/d^2

The Attempt at a Solution

If we choose a point P along the X-axis, with a distance X from the origin, we get this:

$$E_{P1}=\frac{kQ_1}{d^2}$$
$$E_{P2}=\frac{kQ_2}{d^2}$$

Assuming that Q=1 for simplicity:
$$E_{P1}=\frac{k(-7)}{(L+x)^2}$$

and

$$E_{P2}=\frac{k(3)}{x^2}$$

Basically, I've chosen P to be to the right of Q2, giving the distance between Q1 and P to be L+x and the distance between Q2 and P to be x.

If we want to find the point in space where the sums of the Electric fields is zero, we add up the field from Q1 and the field from Q2 and set equal to zero.

$$E_{P1}=\frac{k(-7)}{(L+x)^2} + \frac{k(3)}{x^2} = 0$$

$$E_{P1}=\frac{k(-7)}{(L+x)^2} = - \frac{k(3)}{x^2}$$

I end up with, after assuming L=1, cross-multiplying:

-7x^2 = (-3)(1+x)^2

But when I solve this, there is no real answer.

 

[faradayslaw]

You chose the correct region to look for a net 0 field, (to the right of the 3.00q charge). However, when I solved this equation, I got x==6L.

0==-7k/(L+x)^2+3k/x^2
7/(L+x)^2==3/x^2
7==3(L+x)^2/x^2
7x^2==3(L+x)^2
7/3==(L+x)^2/x^2
7/3==(L^2+2xL+x^2)/x^2
Let L==1
7/3-2==2/x
0.3333333x==2
x==6
Since we let L==1, x==6*L==6

 

[Doc Al]

I end up with, after assuming L=1, cross-multiplying:

-7x^2 = (-3)(1+x)^2

But when I solve this, there is no real answer.

Try again. You should get two real answers, only one of which is meaningful for this problem.

 

[exitwound]

The only two regions is could possibly be zero are the outer two. The inner region would push a positive test charge to the left at all points. But on the outer two, there are possibly points where it could stabilize. How can I possible determine just by looking which one to try?

1.) To the right of the right particle.
If a positive test particle was placed here, it would be repelled by the 3Q charge and attracted by the -7Q charge. But the 3Q charge is much closer.

2.) To the left of the left particle.
If a positive test particle was placed here, it would be repelled by the 3Q charge and attracted by the -7Q charge. But the -7Q charge is much closer.

So just by looking, how can I tell which to choose?

0==-7k/(L+x)^2+3k/x^2
7/(L+x)^2==3/x^2
7==3(L+x)^2/x^2
7x^2==3(L+x)^2
7/3==(L+x)^2/x^2
7/3==(L^2+2xL+x^2)/x^2
Let L==1
7/3-2==2/x <--------//Followed all the way up to here.
0.3333333x==2
x==6
Since we let L==1, x==6*L==6

At the marked point, I have identical work on my paper. How did you get to the next line?

 

[Doc Al]

So just by looking, how can I tell which to choose?

The only way the field contributions can cancel is if you're closer to the smaller charge.

 

[faradayslaw]

The only two regions is could possibly be zero are the outer two. The inner region would push a positive test charge to the left at all points. But on the outer two, there are possibly points where it could stabilize. How can I possible determine just by looking which one to try?

1.) To the right of the right particle.
If a positive test particle was placed here, it would be repelled by the 3Q charge and attracted by the -7Q charge. But the 3Q charge is much closer.

2.) To the left of the left particle.
If a positive test particle was placed here, it would be repelled by the 3Q charge and attracted by the -7Q charge. But the -7Q charge is much closer.

So just by looking, how can I tell which to choose?



At the marked point, I have identical work on my paper. How did you get to the next line?

7/3==2.3333333333333333
2.3333333333333333333-2==0.3333333333333333333
I then brought the x term onto the left side of the equation, 0.333x==2, and divided by 0.333. x==6

 

[Doc Al]

7/3==(L^2+2xL+x^2)/x^2
Let L==1
7/3-2==2/x

No, letting L = 1 gives you:
7/3 = (1 + 2x + x^2)/x^2

(You still have to solve the quadratic.)

 

[exitwound]

Okay, well...

This time I got it down to: 4x^2-6x-3=0

X turns out to be: -.395 and 1.90.

is 1.9 the correct answer? I don't want to lose any more points on this stupid assignment.

Also, why would I disregard the negative one? I know theoretically why i can't have a particle between the two and have a 0 E-field, but is this the reason why I'm disregarding the negative distance?

 

[queenofbabes]

Yes, since you assumed L = 1, x=-.395 will be between the two charges.

 

[exitwound]

1.90 is a wrong submission. Apparently, that is not the answer.

 

[queenofbabes]

You ought to count from the origin...

 

[exitwound]

I don't understand how to relate x to L in this problem.

 

[queenofbabes]

What are the coordinates of q1 and q2? Therefore what is the coordinate of the point with zero E field?

 

[kaymant]

As should be clear by now, the field can be zero only on the right of q₂. Suppose it is zero at a distance s from it. Then, at that point the magnitude of the fields due to the two charge must be same. As such we get (after canceling $$\dfrac{1}{4\pi\epsilon_0}$$ on both side):
$$\dfrac{|q_1|}{(L+s)^2}=\dfrac{|q_2|}{s^2}$$
That, is
$$\dfrac{|q_1|}{|q_2|}=\left(\dfrac{L+s}{s}\right)^2$$
so that
$$\dfrac{L+s}{s}=\pm\sqrt{\dfrac{|q_1|}{|q_2|}}$$
I can readily do away with the negative sign (all the quantities on the left are positive, since I called s as distance). So we get
$$\dfrac{L}{s}+1=\sqrt{\dfrac{|q_1|}{|q_2|}}$$
Finally, I get
$$s=\dfrac{L}{\sqrt{\frac{|q_1|}{|q_2|}}-1}$$
Plugging in numbers,
$$s=\dfrac{L}{\sqrt{7/3}-1}=1.90\,L$$
(taking the significant figures into account)

 

[exitwound]

I'm lost at what I should enter into the submission box. 1.90 is NOT right, as it yelled at me and threw a pie at my face when I tried. Is the problem asking for a submission of 2.90? (If L=1, s=1.9, L+s=2.90)

 

[queenofbabes]

Yes, it should be wanting 2.90.

 

[kaymant]

Yes the answer for the problem would be 2.90. In my last post I found the distance from q₂ of the point where the field is zero. The problem asks for the coordinates, i.e. the distance from the origin which is where q₁ sits. So the answer should be L+s = 2.90 L
So u fill in 2.90.

 

[exitwound]

2.90 is the correct answer.

These problems given on this online homework section are TERRIBLY overcomplicated. Why on Earth is it asking for an answer in multiples of L? I've never seen a question so convoluted in a math book. Why doesn't it just ask straight up "At what point on the x-axis..."??

 

[queenofbabes]

They're really pretty much the same thing isn't it? If L is unknown you will have to give your answer in terms of L anyway.

 

[Doc Al]

I'm lost at what I should enter into the submission box. 1.90 is NOT right, as it yelled at me and threw a pie at my face when I tried. Is the problem asking for a submission of 2.90? (If L=1, s=1.9, L+s=2.90)

Yes. Don't forget that when you set up your equation for x, x was the distance to the right of q2. To get the distance from the origin (where q1 is) you have to add L.

 

[exitwound]

Indeed. But asking, "As a multiple of L...", is extremely confusing.