Is Your Charge Density Integration Correct for a Spherical Volume?

[rudransh verma]

Homework Statement

The volume charge density of a solid non conducting sphere of radius R= 0.056 m varies with radial distance r as given by rho=35.4r/R pC/m^3.
What is the sphere total charge?
What is the field magnitude E at r=0, r=R/3, r=R ?

Relevant Equations

V= 4/3 pir^3
E=1/4pie0q/r^2

First I did drho/dr which is equal to 35.4*10^-12/R. Then I integrated drho by which I got rho=35.4*10^-12. And then the last eqn will be q=rhoV. But the answer was wrong.
I have a doubt on the formula I am using for E because that formula is for a point charge or a charged shell.

 

[mjc123]

First I did drho/dr which is equal to 35.4*10^-12/R. Then I integrated drho by which I got rho=35.4*10^-12.

Why did you differentiate it then integrate again?

And then the last eqn will be q=rhoV.

q is not rho*V because rho is not constant over the volume. You have to integrate ∫rho*dV

 

[kumusta]

Let us see ...

Homework Statement: ...
Relevant Equations: V= 4/3 pir^3 ... E=1/4pie0q/r^2 ..
... I have a doubt on the formula I am using for E because that formula is for a point charge or a charged shell.

... E=1/4pie0q/r^2$~\Rightarrow~E = \frac {1} {4πε_0} \frac {q} {r^2}~\Leftarrow~\begin{cases}\begin{align} & \text {for a point charge}~q~, \text {a mathematical} \nonumber \\ &\text {point without any size nor shape,} \nonumber \\& \text {without any extension in space}~\dots\nonumber \end{align}\end{cases}$
For an extended 3D body (a point large enough to have size and shape in 3D but still small enough to be treated as a point in comparison to other neighboring bodies, the concept of a volume element $~dV$ is used. The total volume $V$ is then the sum of all the volume elements $~dV_i~$ comprising the body: $~V=\sum_{i}$ $dV_i$. If the volume elements are sufficiently small, then they can be treated as point partcles and the total 3D region containing them, including the empty spaces between them, can be said to be occupied by a discrete set of particles.
But if the sufficiently small volume elements are so close together that there are no empty spaces left between them, then the sum is replaced by an integral: $~V=\int_\rm R$ $dV$. The total region R enclosing all the volume elements is then said to be occupied by a continuous distribution of elements.
In the case of a charge $q~$sufficiently small to be treated as a point charge, we have the ordinary product $q = ρV$, where $ρ$ is the charge density. But if $q~$is not small enough and its size and shape must be properly taken into account, then $q~$must be divided into sufficiently, smal enough charge elements $~dq~$, each one occupying the volume element $dV$, whose integral $\int_\rm R$ $dq = q = \int_\rm R$ $ρdV~$gives the total charge $q$. If the charge density is not constant but varies inside the region R, then $ρ$ cannot be taken out of the integral sign.
The preceding is true only for scalar quantities like the total charge. For a vector, like the electric field vector, the integral $\int d\vec E = \vec E$ is more difficult to evaluate directly. Instead, the magnitude and direction are treated separately.
... $dq~\Rightarrow~$production of field $d\vec E~$with scalar magnitude $dE =\frac {1}{4πε_0} \frac {dq}{r^2} = \frac {1} {4πε_0} \frac {ρdV} {r^2}$...
Using the convention that $\vec E$ vectors at the field point move away from positive charges while they move toward negative charges, addition or cancellation of scalar components of vector can be achieved by using symmetry considerations.

 

[rudransh verma]

q is not rho*V because rho is not constant over the volume.

O yes!

You have to integrate ∫rho*dV

Why dV? Are we taking a shell of volume dV with charge dq in it and it’s density as rho. Then we put the value of rho given and 4pir^2dr in place of dV. So the integral is in terms of r. Integration from 0 to .056 m.

 

[mjc123]

Yes

 

[kumusta]

Since $ρ$ is the volume charge density, you have to multiply it by the volume element $dV$.

 

[jbriggs444]

O yes!

Why dV? Are we taking a shell of volume dV with charge dq in it and it’s density as rho. Then we put the value of rho given and 4pir^2dr in place of dV. So the integral is in terms of r. Integration from 0 to .056 m.

It is a shorthand. If you are integrating over a volume, you add up the contributions of a bunch of infinitesimal volume elements. If you are integrating over an area, you add up the contributions of a bunch of infinitesimal area elements. If you are integrating over a line, you add up the contributions of a bunch of infinitesimal line elements.

Rather than go into the possibly messy details of whether one is doing a triple integral with a raster scan of cartesian coordinates x, y and z, a triple integral using polar coordinates, $r$, $\theta$, $\phi$, a single integral over spherical shells, a double integral of rings into a stack of discs or whatever else, one can just say that it is an integral of volume elements over a total volume:: $\int \rho\ dV$.

 

[kumusta]

It is not as simple as you might think. Since the charge distribution is a solid sphere, you have to integrate in three dimensions in the general case. And because of the spherical symmetry of the charge distribution, it is best to use the volume element in spherical coordinate system, again in the general case.

 

[rudransh verma]

Since $ρ$ is the volume charge density, you have to multiply it by the volume element $dV$.

Is dq/dV(or like dv/dt, dx/dt)charge per unit volume like say deltaq/deltaV(delta v/delta t, deltax/delta t).

I hope taking limit doesn’t change the unit of a function delta fx/ delta x.

 

[kumusta]

It's not like the derivative $dx/dt$ since $t$ in the denominator is just a single variable. The volume element $dV$ is like length×width×height where the length, width, and height are all line elements like $dx$ . Have you not studied yet the volume element in spherical coordinate system?

 

[rudransh verma]

It's not like the derivative $dx/dt$ since $t$ in the denominator is just a single variable. The volume element $dV$ is like length×width×height where the length, width, and height are all line elements like $dx$ . Have you not studied yet the volume element in spherical coordinate system?

Ok. But is it(dq/dV)charge per unit volume like delta q/deltaV ,only just more accurate rate?
Or dx/dt or dv/dt are all something per unit something like it’s delta equivalent?

 

[jbriggs444]

Ok. But is it(dq/dV)charge per unit volume like delta q/deltaV ,only just more accurate rate?
Or dx/dt or dv/dt are all something per unit something like it’s delta equivalent?

Yes. You will not go far wrong if you consider dx, dt, dq, dV, etc as infinitesimal elements -- small enough so that there is zero inaccuracy associated with variations in density across the incremental dx, dt, dq or dV.

 

[kumusta]

$dV$ means $V$ is sufficiently small while $\Delta V$ means that $V$ is not small enough to be treated as without having any size or shape.

 

[rudransh verma]

$dV$ means $V$ is sufficiently small while $\Delta V$ means that $V$ is not small enough to be treated as without having any size or shape.

Ya! But first derivative(df/dx)is nothing but rate which comes equal to some algebraic function or numerical. And for different values of x we get different rates at x. Like dx^2/dx=2x which is nothing but rate in terms of x. Limiting makes it more accurate. We say instantaneous rate.
What is the shape of element then?

 

[kumusta]

$df/dx$ means the rate of change of the function $f$ with the variable $x$ . If this rate of change is not constant, then it is just another function $g(x) = df/dx$ . The limiting process makes it more accurate because as the change $\Delta x$ is made smaller and smaller, we get a more accurate value for the change $\Delta f$, until we eventually get the finally value $df/dx$ .

 

[kumusta]

If an element is small enough its actual size and shape do not matter anymore because then it is vanishingly small already.

 

[rudransh verma]

$df/dx$ means the rate of change of the function $f$ with the variable $x$ . If this rate of change is not constant, then it is just another function $g(x) = df/dx$ . The limiting process makes it more accurate because as the change $\Delta x$ is made smaller and smaller, we get a more accurate value for the change $\Delta f$, until we eventually get the finally value $df/dx$ .

Yes! Without limit as we decrease deltax we get more accurate delta fx. But when we take limit we don’t get a more accurate value of fx but we get df/dx at an instant.

 

[kumusta]

$\Delta x$ is reduced without limit because it is impossible to get $\Delta x = 0$. The interval $\Delta x$ is the interval between any two points $x_1$ and $x_2$ and in mathematics we say that there is an infinite number of points between. So it is just impossible to get $\Delta x = 0$. But as we make $\Delta x$ smaller and smaller still, we come to a point where even though $\Delta x$ and its corresponding value $\Delta f$ still change, we always get the same value for $\Delta f/\Delta x$ and that's when we stop taking the limiting process.

 

[rudransh verma]

we always get the same value for Δf/Δx and that's when we stop taking the limiting process.

And that is when delta fx/deltax = df/dx instead of being equivalent .

 

[kumusta]

The limiting process is finally ended when the value for $\Delta f/\Delta x$ combined always comes out the same, even though the separate values for $\Delta f$ and $\Delta x$ still change, because values more accurate than the preceding ones can no longer be obtained. And yes, that is when $\Delta f/\Delta x$ and $df/dx$ become equivalent.

 

[rudransh verma]

The limiting process is finally ended when the value for $\Delta f/\Delta x$ combined always comes out the same, even though the separate values for $\Delta f$ and $\Delta x$ still change, because values more accurate than the preceding ones can no longer be obtained. And yes, that is when $\Delta f/\Delta x$ and $df/dx$ become equivalent.

here 2x+deltax is just another form of deltafx/deltax. We don't know what delta fx separately is here but we know deltax. As we decrease delta x here we will get smaller and smaller (2x+delta x) which will never converge to a value. It only converges when we take deltax=0.
You said deltafx/deltax will stop changing. I don't see it here.

 

[kumusta]

I thought you want to solve a problem in physics involving total charge and the electric field? Now, all what we're talking about is differential calculus. Anyway, it doesn't really matter. If you need help in mathematics, that's okay.
To understand the derivative of a function like the one that you gave, $f(x) = x^2$, we cannot confine ourselves to that only. We have to look more closely to the given function because different functions give different rates of change $\Delta f/\Delta x$.
Now, to know the change $\Delta f$ of a function $f(x)$ , one must first have a change $\Delta x$ in the independent variable $x$ . But without a particular value for $x$, say $x = x_0$ , one cannot talk about its change. But before changing $x_0$ one must first substitute $x = x_0$ in $f(x)$ because one would like to know later how $f_0 = f(x_0)$ changes when another value of $x$ is used. After that, one changes the value of $x_0$ by a certain amount, say $\Delta x$ , so that one now has a new value for $x$ , namely $x_1 = x_0 + \Delta x$. Then, $x_1$ is substituted in $f(x)$ to get its new value $f_1 = f(x_1)$ . Afterwards, examine changes in $\Delta f = f_1 - f_0$ caused by changes in $(x_1 - x_0)$ brought about by slowly reducing the value of
$\Delta x$ in $x_1$ while keeping the value of $x_0$ fixed.Try it for $f(x) = x^2$ with $x_0 = 2$ , $\Delta x = 0.1$ and then reducing $\Delta x$ to the values $\Delta x$ = 0.01 , 0.001 , 0.0001 . Have fun while learning.

 

[rudransh verma]

examine changes in Δf=f1−f0 caused by changes in (x1−x0) brought about by slowly reducing the value of
Δx in x1 while keeping the value of x0 fixed.Try it for f(x)=x2 with x0=2 , Δx=0.1 and then reducing Δx to the values Δx = 0.01 , 0.001 , 0.0001 . Have fun while learning.

It’s coming as 4.1, 4.001, 4.0001 and so on.

 

[kumusta]

Do you understand now what I meant by saying that $\Delta f/\Delta x$ slowly approaches a fixed value and eventually stop changing ?

 

[rudransh verma]

Do you understand now what I meant by saying that $\Delta f/\Delta x$ slowly approaches a fixed value and eventually stop changing ?

But I don’t think it will ever stop at a fixed value.It will come as 4.000000000000000000000000000000000000….and then some digit….

 

[kumusta]

Of course it will since $4.0 = 4.0 × 10^{-10} = 4.0 × 10^{-100} = 4.0 × 10^{-1000} ~\dots~\rm {etc}$ . But if you want the negative exponent of $10^{~n}$ to stop, nobody can ever do it. Exactly the value of 4.0 means $4.0 × 10^{-\infty}$. If you count from 1 to $\infty$, you will never stop counting.

 

[rudransh verma]

Of course it will since $4.0 = 4.0 × 10^{-10} = 4.0 × 10^{-100} = 4.0 × 10^{-1000} ~\dots~\rm {etc}$ . But if you want the negative exponent of $10^{~n}$ to stop, nobody can ever do it. Exactly the value of 4.0 means $4.0 × 10^{-\infty}$. If you count from 1 to $\infty$, you will never stop counting.

So we assume deltaf/deltax value as 4. Because it’s not changing we have given it a name df/dx.
I want to clarify in terms of limits. Taking limit of 2x+deltax , we get values closer and closer to 2x. And we say limit is 2x. And this limit is defined as df/dx. 4 is the rate at x=2.

 

[kumusta]

Yes, its final value is 4.0 since $f(x) = x^2~\Rightarrow~df/dx = 2x$ and evaluating $df/dx$ at $x_0 = 2$ gives $f'(x_0) = 2x_0 = 2(2) = 4 = f'(2)$.

 

[kumusta]

Again, $f(x) = x^2~\Rightarrow~df/dx = 2x$ means that the rate of change of the function $f(x)$ with the independent variable $x$ is given by the function $g(x) = 2x$.
When $x = 2$, the rate of change of the function $f(x)$ is given by $g(2) = 2(2) = 4$.
When $x = 3$, the rate of change of the function $f(x)$ is given by $g(3) = 2(3) = 6$.
We see that as $x$ increases, the rate of change of $f(x)$ with $x$ also increases.

 

[rudransh verma]

We see that as x increases, the rate of change of f(x) with x also increases.

Ok! Suppose this f(x) is f(t)=t^2 a distance function with respect to time t. Then the rate will be 2t. At t=2 sec then velocity will be 4 m/s ?
At t=3 sec velocity will be 6 m/s ?

 

[kumusta]

Ok! Suppose this f(x) is f(t)=t^2 a distance function with respect to time t. Then the rate will be 2t. At t=2 sec then velocity will be 4 m/s ?
At t=3 sec velocity will be 6 m/s ?

$f(t) = t^2\Rightarrow~$a distance function of time $\Rightarrow~x(t) = At^2~\Leftarrow~A = 1~\rm m/s^2 = constant$
$\Rightarrow~dx/dt = 2At = v(t)~\dots$
$\Rightarrow~v(2~{\rm s}) = 2A(2~\rm s) = 2(1~{\rm m⋅s^{-2}})(2~\rm s) = 4~\rm m/s~\dots$
$\Rightarrow~v(3~{\rm s}) = 2A(3~\rm s) = 2(1~{\rm m⋅s^{-2}})(3~\rm s) = 6~\rm m/s~\dots$
I included the constant $A$ with the proper unit so that I get the correct unit for $x$ in meter.

 

[jbriggs444]

But I don’t think it will ever stop at a fixed value.It will come as 4.000000000000000000000000000000000000….and then some digit….

This is not a correct intuition for how limits work.

However, you are correct that it never stops at a fixed value. That is also not how limits work.

So let us work on that. What does it mean when we talk about the limit of a function at a point? In particular, let us use the example at hand. What is the limit of $f(x) = \frac{(4 + 4 \Delta x + (\Delta x)^2) - 4}{\Delta x}$ as $\Delta x$ approaches zero?

You do agree, I hope, that the above limit matches the definition of the derivative of the function $x^2$ at the point 2. We can do a little simplification in the numerator of course so that our starting point is:
$$\lim_{\Delta x \to 0}\frac{4\Delta x + (\Delta x)^2}{\Delta x}$$If we were listening to the teacher, blindly accepting the rules of limits that we are taught, nodding our heads and saying "yes ma'am" we could glance at this limit and immediately say that it is equal to four. But that is not good enough. We want to know why it is exactly 4 and not something like 4.000<infinitely many zeroes>1.

The limit of a function at a point is not the value of the function somewhere very near that point.

The limit of a function at a point is not necessarily the value of the function at the point. [For a continuous function, the value of the function at a point will be equal to the limit of the function at that point].

The limit of a function at a point is the value (if there is one) that is approached ever more closely as one evaluates the function every more closely to the point.

The standard way of expressing this involves epsilons and deltas. One can quickly google up many copies of the definition. For example:

Limit of a function ($\varepsilon-\delta$ definition)

Let $f(x)$ be a function defined on an open interval around $x_0$ ($f(x_0)$ need not be defined). We say that the limit of $f(x)$ as $x$ approaches $x_0$ is $L$, i.e.
$$\lim _{ x \to x_{0} }{f(x) } = L$$
if for every $\varepsilon > 0$ there exists $\delta >0$ such that for all $x$:$$0 < \left| x - x_{0} \right |<\delta \textrm{ } \implies \textrm{ } \left |f(x) - L \right| < \varepsilon$$

[Took a lot of work restoring the embedded $\LaTeX$ in that quote]

This epsilon-delta formulation can be difficult to grasp. Its advantage is that it takes the idea of time and "approaching" out of the picture. The idea of getting closer over time is OK for an intuition, but it is not OK for formal mathematics.

The idea is almost that of a game. You decide how close you want to the formula to get to the limit of 4. That is, you pick an epsilon ($\varepsilon$). I then tell you how close you need to stay to $\Delta x$ = 0. That is, I pick a delta ($\delta$).

If I can always successfully pick a delta ($\delta$) so that value of $\frac{4\Delta x + \Delta x^2}{\Delta x}$ stays within epsilon ($\varepsilon$) of 4 whenever $\Delta x$ is within delta ($\delta$) of zero then the I win and the limit is 4. [We never consider the case where $\Delta x$ is exactly zero]

If you can successfully pick an epsilon ($\varepsilon$) so that I cannot succeed then you win and the limit is not 4 (and may not even exist).

The derivative of x^2 at 2 is exactly 4. It is not an approximation. It is exact.

 

[rudransh verma]

@kumusta we know (2x+deltax) which is deltafx/deltax will never stop at a fixed value as delta x approach 0. So we take limit of 2x+deltax and we assign a value to df/dx as 2x which is a rate we can never have. It’s close to 2x but never 2x.

 

[rudransh verma]

blindly accepting the rules of limits that we are taught, nodding our heads and saying "yes ma'am" we could glance at this limit and immediately say that it is equal to four. But that is not good enough.

Isn’t that applicable to everything we are taught in school and colleges.
And then there comes a time we don’t understand a single word of high level concepts taught in graduation and post graduation!

 

[jbriggs444]

Isn’t that applicable to everything we are taught in school and colleges.
And then there comes a time we don’t understand a single word of high level concepts taught in graduation and post graduation!

That has not been my experience. Certainly there are things that are difficult to grasp. But one can usually get at least a toehold on the edges.

In the case at hand, the teachers are correct. The limit is four exactly. The epsilons and deltas provide a way to define limits in the first place and then to prove this result rigorously, even though they might seem too esoteric to grasp at first.

 

[SammyS]

First I did d(rho)/dr which is equal to 35.4*10^-12/R. Then I integrated d(rho) by which I got rho=35.4*10^-12. And then the last eqn will be q=rho V. But the answer was wrong.
I have a doubt on the formula I am using for E because that formula is for a point charge or a charged shell.

Back to your Original Post. (OP)
You are given the charge density, ρ, as a function of r.

If you differentiate that, with respect to r, and then integrate over r, you get charge density back - in some form. If you find an indefinite integral, and the evaluate the constant of integration, you simply get the charge density back as a function of r. If you evaluate the definite integral from 0 to R, as you did, you simply get the value of the charge density at r = R, i.e., at the surface of the sphere.

To find the charge over some volume you need to do a volume integral, as has been mentioned. In your case, with spherically symmetric charge distribution, the volume element that's handy to use is $dV=4\pi r^2dr$ .

Then $\displaystyle Q_{in}=35.4 (pC/m^3) \int_0^a \frac{r}{R} 4\pi r^2 dr$ is the amount of charge inside a sphere of radius $a$ for $a \le R$ .