Why is the electric field strength not zero at x \gg R?

[Saketh]

A point charge $q$ is located at the center of a thin ring of radius $R$ with uniformly distributed charge $-q$. Find the magnitude of the electric field strength vector at the point lying on the axis of the ring at a distance $x$ from its center, if $x \gg R.$​

I managed to solve the problem to find the electric field strength as a function of x:
$$E(x) = \frac{1}{4\pi \epsilon_0}\left [\frac{q}{x^2} - \frac{qx}{(R^2+x^2)^{3/2}}\right ]$$.

However, I'm having some troubles with the $x \gg R$ part of it. I assumed that this meant that $R \rightarrow 0$, and my function, when R was set to zero, became zero. But the answer says
$$E = \frac{3qR^2}{4\pi \epsilon_0 R}$$
First of all, I don't understand why the R is still there. Second, I don't understand why letting R go to zero is incorrect. If someone could please clarify why the answer is not zero, but is instead this last expression, I would appreciate it.

 

[nrqed]

A point charge $q$ is located at the center of a thin ring of radius $R$ with uniformly distributed charge $-q$. Find the magnitude of the electric field strength vector at the point lying on the axis of the ring at a distance $x$ from its center, if $x \gg R.$​

I managed to solve the problem to find the electric field strength as a function of x:
$$E(x) = \frac{1}{4\pi \epsilon_0}\left [\frac{q}{x^2} - \frac{qx}{(R^2+x^2)^{3/2}}\right ]$$.

However, I'm having some troubles with the $x \gg R$ part of it. I assumed that this meant that $R \rightarrow 0$, and my function, when R was set to zero, became zero. But the answer says
$$E = \frac{3qR^2}{4\pi \epsilon_0 R}$$
First of all, I don't understand why the R is still there. Second, I don't understand why letting R go to zero is incorrect. If someone could please clarify why the answer is not zero, but is instead this last expression, I would appreciate it.

It's a question of approximation.
Ina problem like this, you first try letting R completely negligible compared to x and you get zero, as you mentioned. That tells you that you pushed the approximation a bit too much. Because you know the results cannot be *exactly* zero. The two E fields cancel partially but they can't be cancelling *exactly*.

So what you what to do is to use a Taylor approximation.

Consider $${x \over (R^2 + x^2)^{3/2} }$$
What youmust do is to do a Taylor approximation of that for R <<x.
You write this as
$${1 \over x^2} {1 \over (1 + {R^2 \over x^2})^{3/2} }$$
Now Taylor expand this by treating R/x as your small quantity. This gives
$${1 \over x^2} ( 1 + \ldots)$$

Keeping only the 1 is what you did before but that's not precise enough since it gives a total e field equal to zero (which we know is not correct). So that means that you need the second term in the above expansion. If you include that term, when you will add the two E fields you should get the answer of the book (which is stil an aprroximation since there are other terms in the expansion).

The idea is that you are looking for the first *nonzero* result for the total E field. So if the lowest order approximation (setting R=0 exactly) does not work, you must proceed to the next order.

Hope this makes sense.

 

[-marko-]

I attached my solution in pdf format. Similar solution you can find on this page http://irodov.nm.ru/3/resh/3_10.gif

 

[nrqed]

I attached my solution in pdf format. Similar solution you can find on this page http://irodov.nm.ru/3/resh/3_10.gif

I haven't seenyour solution yet but please know that it is highly discouraged to post complete solutions to problems. The goal is not to provide solutions but to help people work out through their problems by giving hints and some steps. The goal is to help people learn, not do their problems for them. Of course, people *may* learn by seeing a complete solution, but that is not the most pedagogical way to get people to understand the physics and maths.

Regards,

Patrick

 

[Saketh]

Okay, I understand now. I never knew that you were supposed to use Taylor expansions for this sort of approximation.

But now I know.

 

[-marko-]

I haven't seenyour solution yet but please know that it is highly discouraged to post complete solutions to problems. The goal is not to provide solutions but to help people work out through their problems by giving hints and some steps. The goal is to help people learn, not do their problems for them. Of course, people *may* learn by seeing a complete solution, but that is not the most pedagogical way to get people to understand the physics and maths.

Regards,

Patrick

You are absolutely right so please accept my apology.

Regards,

Marko