Why Does Adding an Opposite Point Charge Affect Electric Field Strength?

In summary, His solution to part ii was to take r to be the distance between the point charges and to use the following equation to calculate the field strength: E=2(kQ/(r/2)^2) due to the fact that you are measuring the field strength half way between the point charges.
  • #1
Millsworth
3
0
I am having difficulty explaining this question to my student.

l.jpg


His solution to part ii was (taking r to be the distance between the point charges) E = 2(kQ/(r/2)^2 due to the fact that you are measuring the field strength half way between the point charges.

So, E = 2 x (9E9 x 1.6E-19/(1E-10)^2) = 2.9E11 N/C

The solution given is

l.jpg


This is the solution that is also given in the revision guide.

He cannot understand why adding an opposite point charge which provides a force in the same direction would mean the field strength goes down.


I can't give him an answer that satisfies. Can anyone help?
 
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  • #2
Millsworth said:
He cannot understand why adding an opposite point charge which provides a force in the same direction would mean the field strength goes down.

The field strength at that point will go up, not down. (compared to the field strength at that point if only the positive charge was present.) Also, the distance from each charge to the point directly between them is 1*10^-10m not 2*10^-10m.

If I am reading the question correctly, I agree with your students answer:

There are two charges: [itex]Q_1=1.6\times10^{-19}C[/itex] [itex]Q_2=-1.6\times10^{-19}[/itex] separated by 2*10^-10m.

Therefore the point halfway between them is located 1*10^-10m away from each charge. Therefore [itex]r_1=r_2=r=1\times10^{-10}.[/itex]

The fields from each charge point in the same direction, so they add:

[tex]E=E_1+E_2=KQ_1/r_1^2 +KQ_2/r_2^2= \frac{2KQ}{r^2}[/tex]
 
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  • #3
Thank you for your swift response

Are you saying that you think the exam solution is wrong?

Is my students answer correct?
Should the solution to ii be 2(kQ/(r/2)^2?
 
  • #4
If only the positive charge was present E = 1.4x10^11 N/C

If opposite charge is added, OCR say E = 7.2X10^10 N/C
 
  • #5
Millsworth said:
Is my students answer correct?
Should the solution to ii be 2(kQ/(r/2)^2?

Yes.

I've edited my above response to show my reasoning. Sorry it took so long. I had problems with LaTeX.
 
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  • #6
Millsworth said:
If only the positive charge was present E = 1.4x10^11 N/C

This is correct.
If opposite charge is added, OCR say E = 7.2X10^10 N/C

This is wrong. Think about it. We are looking for the field at the center point. So, how far away from the center point is each charge? That is r.

[itex]r \not= 2\times10^{-10}[/itex]

[itex]r=1\times10^{-10}[/itex]
 
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  • #7
If you label the separation of A and B as r, then the field midway between them is:

[tex]E=\frac{kq_A}{(\frac{r}{2})^2} + \frac{kq_B}{(\frac{r}{2})^2}[/tex]

Since qA=-qB, I'll call the magnitude of their charge q:

[tex]E=\frac{kq}{(\frac{r}{2})^2} - \frac{-kq}{(\frac{r}{2})^2} = \frac{2kq}{(\frac{r}{2})^2}[/tex]Using the given values of q and r:

[tex]E= \frac{2kq}{(\frac{r}{2})^2} = \frac{2(9\times 10^9~Nm^2C^{-2})(1.6\times 10^{-19}~C)}{(1\times 10^{-10}~ m)^2}=2.88\times 10^{11}~N/C[/tex]So your student was correct.
 

1. What is an electric field?

An electric field is a physical quantity that describes the influence that a charged object has on other charged objects in its surroundings. It is a vector field, meaning it has both magnitude and direction, and is created by the presence of electric charges.

2. What is the difference between an electric field and an electric potential?

An electric field is a measure of the force that a charged object would experience at a given point, while electric potential is a measure of the work required to move a unit charge from one point to another. In other words, electric potential is the potential energy per unit charge at a given point in an electric field.

3. How do you calculate the strength of an electric field?

The strength of an electric field is calculated by dividing the force experienced by a test charge by the magnitude of the test charge. In mathematical terms, it is represented by the equation E = F/Q, where E is the electric field strength, F is the force, and Q is the magnitude of the test charge.

4. What factors affect the strength of an electric field?

The strength of an electric field is affected by the magnitude of the source charge, the distance from the charge, and the medium in which the field exists. The strength decreases as the distance from the source increases and also depends on the medium's permittivity, which describes its ability to store electric charge.

5. How is an electric field represented visually?

An electric field is represented visually by using electric field lines. These lines indicate the direction of the electric field at different points in space and are drawn perpendicular to the surface of a charged object. The closer the lines are to each other, the stronger the electric field at that point.

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