Finding the Total Electric Field at X=0 on the X-Axis

In summary: No problem! Just remember to always draw a picture and simplify the problem before starting to solve it. That will make it easier to see what the unknowns are and how they relate to each other.In summary, two charges of -11.8 micro-Coulombs and -7.3 micro-Coulombs are placed on the x-axis at x=0 meters and x=0.42 meters respectively. The total electric field on the x-axis, other than infinity, is equal to zero at a distance of approximately <some number> centimeters from the origin, where the ratio of the distance from the origin to the equilibrium point and the distance from the second charge to the equilibrium point is equal to the square root of the ratio of
  • #1
eestep
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Homework Statement


A charge of -11.8 micro-Coulombs is placed on x-axis at x=0 meters. Another charge of -7.3 micro-Coulombs is placed on x-axis at x=+.42 meters. Other than infinity, where on x-axis is total electric field equal to zero? Answer in centimeters.
q1=-11.8[tex]\mu[/tex]C
q2=-7.3[tex]\mu[/tex]C

Homework Equations


E=kq/r2[tex]\widehat{}r[/tex]

The Attempt at a Solution


E1=E1[tex]\widehat{}x[/tex]=-(8.99*1099Nm2/C2)(11.8*10-6C)/r12[tex]\widehat{}x[/tex]
E2=E2[tex]\widehat{}x[/tex]=(8.99*109Nm2/C2)(7.3*10-6C)/r22
Ex=E1x+E2x=-(8.99*109Nm2/C2)(11.8*10-6C)/r12+(8.99*109Nm2/C2)(7.3*10-6C)/r22=0
11.8/7.3=r12/r2^2
r1/r2=[tex]\sqrt{}11.8/7.3[/tex]
 
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  • #2
I didn't check all of your steps. But if that is right then what you want now is to just have the x-coordinate. So really you only want one 'r', the other one can be written in terms of this r as say, r2 = 0.42 - r1.
Then you can just solve for your r which is the x-coordinate of the equilibrium point.
 
  • #3
How do I do that? I get r2^2=7.3r1^2/11.8.
 
  • #4
Well if the ratio is correct:
[tex]\frac{r_{1}}{r_{2}} = \frac{\sqrt{11.8}}{7.3}[/tex]
Just look at the problem in question. Your first charge is at x=0 the second charge is at x=0.42
Now you are asked to find the equilibrium point that should just be a single coordinate. So from the origin to the equilibrium point is r let's say, this also happens to be the distance from the first charge as well since the first charge happens to be at the origin which you called [itex]r_{1}[/itex]. So you have a relation between the distance from the first charge to the equilibrium point and the coordinate of the equilibrium point... which is just: [itex]r_{1} = r[/itex].
Now you want to express the distance from the second charge to the equilibrium point. It should be obvious that it is: [itex]r_{2} = 0.42 - r[/itex] which is the same as: [itex]r_{2} = 0.42 - r_{1}[/itex]

If it's not obvious then draw a picture. Put the equilibrium point between the two charges and come up with those expressions for the distances to each charge, where r is the unknown x coordinate of the equilibrium point.

So therefore your ratio becomes:
[tex]\frac{r}{0.42 - r} = \frac{\sqrt{11.8}}{7.3}[/tex]
Then you just do some algebra and solve for r

you should get r = <some number> that does not depend on [itex]r_{1}[/itex] or [itex]r_{2}[/itex].
 
  • #5
I appreciate your aid!
 

1. What is the formula for finding the total electric field at X=0 on the X-axis?

The formula for finding the total electric field at X=0 on the X-axis is given by the superposition principle, which states that the total electric field at a point is the vector sum of all individual electric fields at that point. Mathematically, it can be expressed as E = ∑(kq/r^2), where E is the total electric field, k is the Coulomb's constant, q is the charge, and r is the distance between the point and the charge.

2. How do you determine the direction of the total electric field at X=0 on the X-axis?

The direction of the total electric field at X=0 on the X-axis is determined by combining the direction of each individual electric field using vector addition. If the individual electric fields are in the same direction, the total electric field will also be in that direction. If the individual electric fields are in opposite directions, the total electric field will be the difference between the two directions.

3. What factors affect the magnitude of the total electric field at X=0 on the X-axis?

The magnitude of the total electric field at X=0 on the X-axis is affected by the magnitude of the individual electric fields, the distance between the charges, and the angle at which the individual electric fields are directed towards the point. The greater the magnitude of the individual electric fields and the smaller the distance between the charges, the higher the magnitude of the total electric field. The angle at which the individual electric fields are directed also plays a role in determining the magnitude of the total electric field.

4. Can the total electric field at X=0 on the X-axis be negative?

Yes, the total electric field at X=0 on the X-axis can be negative. This means that the direction of the total electric field is opposite to the direction of the individual electric fields. This can happen when the individual electric fields have different magnitudes and are directed at different angles towards the point, resulting in a net electric field in the opposite direction.

5. How can you use the total electric field at X=0 on the X-axis in practical applications?

The total electric field at X=0 on the X-axis is an important concept in understanding the behavior of electric charges. It is used in various practical applications, such as in the design of electronic circuits, determining the behavior of charged particles in an electric field, and understanding the properties of materials. It is also used in the calculation of the force exerted on a charge in an electric field, which is important in fields such as electromagnetism and electronics.

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