Where Is the Electric Field Zero in This Dipole Setup?

[morrobay]

Homework Statement

Consider a dipole:
a=50 cm
q1 = +2q on x axis
q2 = -5q on x axis
Point on x-axis where E=0,

Homework Equations

+q1/4pi(eo)x^2 = -q2/4pi(eo)(a-x)^2,

The Attempt at a Solution

algebra =
x= 50/1+ (5/2)^1/2
x = 19.37 cm from +2q
this checks as the magnitudes of
=+2q/19.37^2 and -5q/30.63^2 are equal
The question i have :
In dipole diagrams the E field lines of force show a continuous field from
+q to -q
This appears to be in conflict with the above, that is, opposing E fields where E = 0
at 19.37 cm from +2q on x-axis ?






]

 

[alphysicist]

Hi morrobay,

Homework Statement

Consider a dipole:
a=50 cm
q1 = +2q on x axis
q2 = -5q on x axis
Point on x-axis where E=0,

Homework Equations

+q1/4pi(eo)x^2 = -q2/4pi(eo)(a-x)^2,

The Attempt at a Solution

algebra =
x= 50/1+ (5/2)^1/2
x = 19.37 cm from +2q
this checks as the magnitudes of
=+2q/19.37^2 and -5q/30.63^2 are equal
The question i have :
In dipole diagrams the E field lines of force show a continuous field from
+q to -q
This appears to be in conflict with the above, that is, opposing E fields where E = 0
at 19.37 cm from +2q on x-axis ?

I believe a dipole is always defined to be made of two charges that are opposite in sign but equal in magnitude, so this would not be a dipole.

However, about the question, I don't believe your answer is correct, because you have made an assumption about the answer that is not correct. On a diagram, let q1 be on the left and q2 be on the right. Now at the point you found that is between the charges, and 19cm to the right of q1, what is the direction of the electric field from q1? what is the direction of the electric field of q2? Can you see that it is not possible for them to cancel there?

So on the diagram, where does the point need to be? That will help you rewrite your equations.

(There is also a mathematical way that would get the answer using the equations in your post, but you would still need to decide where to put the charge at based on physical ideas.)

 

[morrobay]

thanks for fast reply.
this would be the diagram

+2q--------------------50cm------------------(-5q)

E=0 at 19.37 cm to right of +2q

My same question would apply to this diagram.
If the following is correct:
1) the E field from +2q is directed to the right
2) the E field from -5q is directed to the left
3) a +q test charge released at +2q would move to -5q

 

[LowlyPion]

thanks for fast reply.
this would be the diagram

+2q--------------------50cm------------------(-5q)

E=0 at 19.37 cm to right of +2q

My same question would apply to this diagram.
If the following is correct:
1) the E field from +2q is directed to the right
2) the E field from -5q is directed to the left
3) a +q test charge released at +2q would move to -5q

Doesn't that suggest then that the point of the field strength = 0 doesn't lie between the charges?

 

[alphysicist]

thanks for fast reply.
this would be the diagram

+2q--------------------50cm------------------(-5q)

E=0 at 19.37 cm to right of +2q

No, what I was saying was that E is not zero 19.37 cm to the right of the +2q charge.

My same question would apply to this diagram.
If the following is correct:
1) the E field from +2q is directed to the right

This is correct, because the field from a positive charge always points away from itself.

2) the E field from -5q is directed to the left

But this one is not correct, because the field of a negative charge always points toward itself.

3) a +q test charge released at +2q would move to -5q

That's true; and more importantly, a test charge (positive or negative) placed at 19.37 cm to the right of the +2q would feel a force. (If it's a positive test charge, it's being pushed to the right by the positive charge, and pulled to the right by the negative charge. If it's negative, those are both reversed.) That shows that the field cannot be zero there.

What I would suggest when doing these problems is to first use your diagram to decide in what region the field can be zero at some point: is it between the charges, to the left of the charges, or to the right of the charges? Once you figure that part out, I think the equations are easier to write down.

 

[morrobay]

The diagram in post #5 and the following question are from:
Physics combined edition, first and second edition Halliday Resnick
chapter 27, page 590 problem #12 ( no answere in back)

"In fig 27-18 locate the point (or points ) at which the electric field strengh is zero take a =50cm."

From the diagram and the question it suggested to me that E=0 at some point on the line between the charges where the opposing E fields cancel each other out.

 

[alphysicist]

The diagram in post #5 and the following question are from:
Physics combined edition, first and second edition Halliday Resnick
chapter 27, page 590 problem #12 ( no answere in back)

"In fig 27-18 locate the point (or points ) at which the electric field strengh is zero take a =50cm."

From the diagram and the question it suggested to me that E=0 at some point on the line between the charges where the opposing E fields cancel each other out.

But that's the problem: between the charges the E fields are not opposing each other! Between the charges both E fields are pointing towards the right, and so they cannot cancel there.

What you first have to do is find out where the E fields are opposing each other, and as I said in my last post there are three regions to check: to the left of the charges, between the charges, and to the right of the charges. (We've already found that the E fields from both charges are to the right between the charges, and so you have to check if the E fields are in opposite directions in the other two regions.)

Once you have answered that, you can narrow it down more by answering this question: Is the point where the E fields cancel going to be closer to q1 or q2? That question can be answered by considering Coulomb's law, and its answer will determine the single region that is possible. What do you get?

 

[morrobay]

Alright i have it now, 86 cm to left of +2q .
The equality should have been q1/x^2=q2/(a+x)^2

 

[LowlyPion]

Alright i have it now, 86 cm to left of +2q .
The equality should have been q1/x^2=q2/(a+x)^2

Yes the equation looks right, but I didn't do the math. It looks about right. A quick check shows 2/ (86)^2 = 5/ (136)^2 ... so I'd say that should do it.