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

[Millsworth]

I am having difficulty explaining this question to my student.

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

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?

 

[G01]

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: $Q_1=1.6\times10^{-19}C$ $Q_2=-1.6\times10^{-19}$ separated by 2*10^-10m.

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

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

$$E=E_1+E_2=KQ_1/r_1^2 +KQ_2/r_2^2= \frac{2KQ}{r^2}$$

 

[Millsworth]

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?

 

[Millsworth]

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

 

[G01]

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.

 

[G01]

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.

$r \not= 2\times10^{-10}$

$r=1\times10^{-10}$

 

[Superstring]

If you label the separation of A and B as r, then the field midway between them is:

$$E=\frac{kq_A}{(\frac{r}{2})^2} + \frac{kq_B}{(\frac{r}{2})^2}$$

Since q_A=-q_B, I'll call the magnitude of their charge q:

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

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