Integration of an equation relating to electrostatics

[Lazy Rat]

Homework Statement

Hi I was wondering if anyone could give me a hand with this problem I'm trying to solve.

I am trying to integrate this equation twice. I'm not really sure what to do with the right hand side of the equation.

Homework Equations

The Attempt at a Solution

[/B]
The left side I am ok with i think,

dV/dr = A/r

Any help as to how to deal with the right hand side would be great.

Thank you

 

[haruspex]

dV/dr = A/r

Is that supposed to be the integral of the left hand side? I don't understand.

how to deal with the right hand side would be great.

How do the variables on the right depend on r?

 

[Abhishek kumar]

Is that supposed to be the integral of the left hand side? I don't understand.

How do the variables on the right depend on r?

Homework Statement

Hi I was wondering if anyone could give me a hand with this problem I'm trying to solve.

I am trying to integrate this equation twice. I'm not really sure what to do with the right hand side of the equation.

Homework Equations

View attachment 215481

The Attempt at a Solution

[/B]
The left side I am ok with i think,

dV/dr = A/r

Any help as to how to deal with the right hand side would be great.

Thank you

This is 2nd order nonhomogeneous differential equation.above equation can be written as
V''+V'(1/r^2)=-ρf/εε°r

 

[Delta2]

if we suppose that ρ,f are constants or known functions $\rho(r),f(r)$, then the 2nd order linear nonhomogeneous and with no constant coefficients ODE I get is , by expanding the left hand side:
$V''+\frac{1}{r}V'=-\frac{\rho f}{\epsilon\epsilon_0}$.

 

[Abhishek kumar]

if we suppose that ρ,f are constants or known functions $\rho(r),f(r)$, then the 2nd order linear nonhomogeneous and with no constant coefficients ODE I get is , by expanding the left hand side:
$V''+\frac{1}{r}V'=-\frac{\rho f}{\epsilon\epsilon_0}$.

Yes your's equation is right i misscalculated

 

[rude man]

The equation can be solved by a reduction of order if, as post 4 states, ρ = ρ(r) and f = f(r) but not functions of V.

 

[haruspex]

if we suppose that ρ,f are constants or known functions $\rho(r),f(r)$, then the 2nd order linear nonhomogeneous and with no constant coefficients ODE I get is , by expanding the left hand side:
$V''+\frac{1}{r}V'=-\frac{\rho f}{\epsilon\epsilon_0}$.

Yes, but as rude man points out, if the variables on the right depend on r, not V, that is a retrograde step.

 

[Lazy Rat]

$-ρf / εε_0$ relates to Poisson equation $pf$ is the free charge density and $ε_0$ is the permitivity of free space. So these can be taken as know constants on this occasion.
I am looking to achieve a general solution through integrating twice. With a similar equation for electrostatics and Laplace's equation I have ${1/r} d/dr(r{dV/dr})=0$ , then integrating once we have ${dV} /{dr} = {A/r}$ then twice $V(r) = A {ln} r + B$ .
So how can I solve this in a similar way regarding Poisson's equation?

Thanks for you input guys

 

[haruspex]

how can I solve this in a similar way

In exactly the same way. Multiply out and integrate. What do you get?

 

[rude man]

BTW the given equation is the poisson eq. for cylindrical coordinates with no variation in θ or z.

 

[rude man]

@Lazy Rat, BTW the expression for ∇²V = -ρ/ε can also be compacted in spherical coordinates to
(1/r²) d/dr {r²dV/dr} = -ρ/ε
with solution proceeding in the same way haruspex pointed to in his post 9.

 

[Lazy Rat]

In exactly the same way. Multiply out and integrate. What do you get?

So I get the same because the right side I treat as a constant. The answer is still the Laplace general solution $V(r)= A~ ln (r)~ + B$
Is this correct?

@Lazy Rat, BTW the expression for ∇²V = -ρ/ε can also be compacted in spherical coordinates to
(1/r²) d/dr {r²dV/dr} = -ρ/ε
with solution proceeding in the same way haruspex pointed to in his post 9.

Thank you rude man, may try to tackle in spherical.

 

[rude man]

So I get the same because the right side I treat as a constant. The answer is still the Laplace general solution $V(r)= A~ ln (r)~ + B$
Is this correct?

The right side IS a constant but when you integrate (twice in fact) that constant has an impact on the answer.
Do the solution step-by-step & we can show you where you went wrong.

 

[Lazy Rat]

Hi
Ok so first step is to multiply out.
Thus we have for $\frac 1r \frac d{dr} (r \frac {dV}{dr})= \frac {\rho f} {\epsilon \epsilon_0}$
$\frac 1r \times r + \frac 1r \times \frac {dV}{dr} + \frac d{dr} \times r +\frac d{dr} \times \frac {dV}{dr} =\frac {\rho f} {\epsilon \epsilon_0}$
$= 1+ \frac {dV}{dr^2} + \frac {dr}{dr} + \frac {d^2V}{dr} =\frac {\rho f} {\epsilon \epsilon_0}$
$= \frac {d^2V}{dr2} + \frac 1r \frac {dV}{dr} = \frac {\rho f} {\epsilon \epsilon_0}$
Is this correct for multiplying out the left side?
Thank you

 

[haruspex]

Is this correct for multiplying out?

No. If you have an equation like y/x=z then multiplying out gives y=xz.

 

[Lazy Rat]

I see, I've epand the brackets not multiplied out.
$\frac d{dr} (r \frac {dV}{dr})= \frac {\rho f} {\epsilon \epsilon_0} r$
Is this correct?
Can the derivative terms such as $\frac d{dr}$ be manipulated algebraically, and why is it $\frac d{dr}$ and not $\frac {dV}{dr}$ are these the same thing ?

All this is a dark room to me, you are shedding light, I appreciate your time. Thank you.

 

[haruspex]

I see, I've epand the brackets not multiplied out.
$\frac d{dr} (r \frac {dV}{dr})= \frac {\rho f} {\epsilon \epsilon_0} r$
Is this correct?.

Yes.
You can immediately integrate that. Don't be distracted by the $(r \frac {dV}{dr})$ term. At this stage it is just some function of r. Think of it as $\frac d{dr} F(r)= \frac {\rho f} {\epsilon \epsilon_0} r$

 

[rude man]

Hi
Ok so first step is to multiply out.
Thus we have for $\frac 1r \frac d{dr} (r \frac {dV}{dr})= \frac {\rho f} {\epsilon \epsilon_0}$
$\frac 1r \times r + \frac 1r \times \frac {dV}{dr} + \frac d{dr} \times r +\frac d{dr} \times \frac {dV}{dr} =\frac {\rho f} {\epsilon \epsilon_0}$
$= 1+ \frac {dV}{dr^2} + \frac {dr}{dr} + \frac {d^2V}{dr} =\frac {\rho f} {\epsilon \epsilon_0}$
$= \frac {d^2V}{dr2} + \frac 1r \frac {dV}{dr} = \frac {\rho f} {\epsilon \epsilon_0}$
Is this correct for multiplying out the left side?
Thank you

Other than you need a - sign in front of $\frac {\rho f} {\epsilon \epsilon_0}$ yes, that is in fact the expanded version for the Poisson equation for cylindrical coordinates.

 

[haruspex]

that is in fact the expanded version

Yes, but it is not "multiplying out", and it is not on the simplest path to solving the equation.

 

[rude man]

Other than you need a - sign in front of $\frac {\rho f} {\epsilon \epsilon_0}$ yes, that is in fact the expanded version for the Poisson equation for cylindrical coordinates.

You previously solved the Laplace equation without this multiplying-out. Stick to what you were doing before, which was double integrating (1/r)d/dr{rdV/dr)} = 0. Just change "0" to "-ρ/ε". You'll get the same first two terms you did with the "0" plus you'll get a third due to the "-ρ/ε".

 

[haruspex]

You previously solved the Laplace equation without this multiplying-out. Stick to what you were doing before, which was double integrating (1/r)d/dr{rdV/dr) = 0. Just change "0" to "-ρ/ε". You'll get the same first two terms you did with the "0" plus you'll get a third due to the "-ρ/ε".

We seem to be offering conflicting advice, but it might just be a question of terminology. I've started a conversation on which we can resolve this, one hopes.

 

[haruspex]

We seem to be offering conflicting advice, but it might just be a question of terminology. I've started a conversation on which we can resolve this, one hopes.

Having discussed it offline, it seems there was a misunderstanding. rude man agrees that the way is to multiply out in the sense of the step taken in post #16, not to expand the derivatives as in post #14.

 

[Lazy Rat]

Ok thanks for clearing up confusion guys

I think I may have got there, I have after multiplying out

$∫ \frac d {dr}( r \frac {dV}{dr}) = ∫ - \frac ρ ε {r}$

Integrating twice

$v(r) = A ~ ln (r) - \frac {ρ r^2} {4 ε} + B$

Is this correct?

Thank you

 

[rude man]

Ok thanks for clearing up confusion guys

I think I may have got there, I have after multiplying out

$∫ \frac d {dr}( r \frac {dV}{dr}) = ∫ - \frac ρ ε {r}$

Integrating twice

$v(r) = A ~ ln (r) - \frac {ρ r^2} {4 ε} + B$

Is this correct?

Thank you

Got it! But you have to append "dr" to both sides of your first equation & clean up the left side.

 

[Lazy Rat]

Got it! But you have to append "dr" to both sides of your first equation & clean up the left side.

Ah yes of course the dr, by clean up the left side you mean don't have the $\frac d {dr}$ in the equation?

Thanks rude man

 

[rude man]

Ah yes of course the dr, by clean up the left side you mean don't have the $\frac d {dr}$ in the equation?

Thanks rude man

When you add "dr" to the left side, two "dr's" cancel, leaving you with
∫d{r dV/dr} = r dV/dr + constant etc.
Like ∫dy = y + constant.

 

[Lazy Rat]

Wonderful, Thank you.