Problem Solving Strategies

sanya

07:09

Hi sir @JohnRennie what does "space charge" mean

John Rennie

Hi :-)

Can you post the question so I can see the context?

sanya

Yes

John Rennie

I think it just means there is some charge density in between the plates that is a function of the position in space.

i.e. we have some charge density ρ(x) in between the plates where x is the position between the plates.

Is that a JEE question? It seems to beyond JEE level ...

sanya

Oh

@JohnRennie It is in one of the jee books examples

Im a little confused it was written somehwere that charge doesnt accumulate at different points when theres a current..?

John Rennie

That is true in a wire, but this is a bit different as we have electrons in a vacuum i.e. it's like an electron "gas" in between the two plates.

So the number of electrons per m³ can vary in the vacuum.

In a wire the density of conduction electrons is the same as the density of the nuclei because the wire is overall neutral.

I can explain how to do this, but it seems more advanced that I would expect for the JEE.

sanya

07:20

Maybe you can try explaining? If I dont get it i'll skip this question. .

John Rennie

You need to know how the potential is related to the charge density, and this is given by Poisson's equation:
∇²V = -ρ/ε₀

Where V is the potential and ρ is the charge density.

In this case the charge density only varies with the distance between the plates, 𝑥, so the ∇²V operator is just d²V/dx²

So the algebra is actually very simple. Just take the V(x) that you're given, differentiate it twice and that gives you ρ(x)/ε₀

d²V/dx² = -ρ(x)/ε₀

But I didn't think Poisson's equation was covered in the JEE.

sanya

The book takes put dv/dx which is E and then uses gauss law tp get the charge density

John Rennie

Ah, OK, yes I guess that would work.

I guess it finds the charge between x and x+dx?

sanya

Yes

John Rennie

OK, that is basically using Poisson's equation because you'll find the net flux is related to dE i.e. the change in the field between x and x+dx, and than you'll get a dE/dx term, which is d²V/dx².

We can go through it if you want...

sanya

07:33

Can we go through the part from after taking out the electric field?

The gauss law bit

John Rennie

OK, give me a moment and I'll draw a diagram ...

We have two plates and some charge density in between them that varies with x.

sanya

Yes

John Rennie

The shading is just meant to indicate the charge density varies - don't take it too literally.

sanya

Okay

John Rennie

We are told V(x) = ax^4/3

And E = -dV/dx

So E = -⁴⁄₃ax^1/3

sanya

07:40

Yes

John Rennie

Now consider the volume enclosed by the two dashed lines.

Let's assume the plates have a unit area so the volume of this slab is just dx.

OK so far?

sanya

Yes

John Rennie

Then the charge inside this slab is dQ = ρ(x) dx

sanya

Yes

John Rennie

Now, the field at the left side is E(x) and the field at the right side is E(x+dx)

sanya

07:43

Yes

John Rennie

And the areas are unity (we assumed area = 1) so the fluxes on the two sides are also just E(x) and E(x+dx). Since the fields point in the same direction the net flux is the difference between these i.e.
Flux = E(x+dx) - E(x)

OK so far?

sanya

Yes

John Rennie

And from Gauss's law the net flux has to be equal to the charge inside the slab divided by ε₀, and from above we found this charge was dQ = ρ(x) dx

Yes?

sanya

Yes

John Rennie

So putting this all together we get:
E(x+dx) - E(x) = ρ(x) dx / ε₀

And we divide by dx to get:
(E(x+dx) - E(x))/dx = ρ(x)/ε₀

Does this make sense so far?

sanya

07:48

Yes

John Rennie

But look at the left hand side. That's just the expression for dE/dx. Yes?

sanya

Oh

Yss

Yes

John Rennie

So we got:
dE/dx = ρ(x)/ε₀

And E = -dV/dx so dE/dx = -d²V/dx²

sanya

Oh which means d²v/d x²=p(x)/epsilon

Yes

John Rennie

Yes, it's simpler than you thought :-)

(you missed the minus sign)

sanya

07:51

Oh yup

John Rennie

-d²V/dx² = ρ(x)/ε₀

That's just Poisson's equation.

sanya

Got it !

John Rennie

(in 1D)

@sanya OK :-)

And the rest is just algebra ...

sanya

@JohnRennie yes

Binky McSquigglebottom

@JohnRennie Hi !!!

John Rennie

08:00

Hi :-)

Binky McSquigglebottom

How do I write the equations of parabolic motion as if there were friction in the fluid?

John Rennie

Do you mean where the drag force is proportional to v²?

Binky McSquigglebottom

I mean for example Ff= -bv

John Rennie

That's linear drag.
Are you considering motion in 1D i.e. along a straight line?

Binky McSquigglebottom

Isn't parabolic motion 2-dimensional?

I'm not understanding

John Rennie

08:09

You mean like throwing a stone at an angle so it moves horizontally as well as vertically?

Binky McSquigglebottom

Yes , i mean projectile motion

John Rennie

There isn't a simple solution. I can show you how to set up the equations, but there is no simple way to solve them.

Binky McSquigglebottom

Okay

John Rennie

Suppose we start with no drag force.

Then we can write equations for the x and y components of the acceleration:
ay = -g
ax = 0
Yes?

Binky McSquigglebottom

Yes

John Rennie

08:14

And converting these to differential equations we get:
d²y/dt² = -g
d²x/dt² = 0

And these two equations can be solved separately because d²y/dt² does not depend on x and d²x/dt² does not depend on y.

OK so far?

Binky McSquigglebottom

OK

John Rennie

But suppose we have a drag force F = -av for some constant 𝑎.

Binky McSquigglebottom

$\alpha$ Is the a in the formula, right ?

Which would be the viscosity coefficient ?

John Rennie

We need to work out the x and y components of the force F because then we can put them in our equations like this:
m d²y/dt² = -mg + Fy
d²x/dt² = Fx
Yes?

Binky McSquigglebottom

I don't understand how you arrived at these formulas

John Rennie

08:22

Consider the y equation. This comes from the second law F = ma. In the absence of drag the force F is the gravitational force -mg (- because it's downwards) and the acceleration 𝑎 is d²y/dt².

So if we substitute these into the second law:
ma = F
we get:
m d²y/dt² = -mg

Yes?

Binky McSquigglebottom

Yes , and long x?

John Rennie

Let's just consider y for now...

Binky McSquigglebottom

OK

John Rennie

If we have drag then we have some drag force F, and we can write this as a y component, Fy, and an x component, Fx.

So now the total force in the y direction is the gravitational force -mg plus the drag force Fy.
Yes?

Binky McSquigglebottom

I don't understand why it is +Fy and not -Fy

John Rennie

08:29

The minus sign will appear when we write F = -αv

Binky McSquigglebottom

Ah ok!

John Rennie

Now, we know F = -αv

And v = √(vx² + vy²)

So F = -α√(vx² + vy²)

Yes?

Binky McSquigglebottom

Yes

John Rennie

And the vertical component of the force is Fy = F sinθ

Yes?

Binky McSquigglebottom

Yes

John Rennie

08:37

So our equation:
m d²y/dt² = -mg + Fy
becomes:
m d²y/dt² = -mg - α√(vx² + vy²) sinθ

OK so far?

Binky McSquigglebottom

Would that be the case?

@JohnRennie ok

John Rennie

@BinkyMcSquigglebottom I've redrawn your diagram to make it clearer.

Binky McSquigglebottom

👍

John Rennie

m d²y/dt² = -mg - α√(vx² + vy²) sinθ

In fact this isn't as complicated as it looks because we can use vy = v sinθ to get an expression for sinθ:
sinθ = vy/v = vy/√(vx² + vy²)

Yes?

Binky McSquigglebottom

But isn't theta already known?

John Rennie

08:46

θ changes as the particle moves along the parabola

It isn't the launch angle.

Binky McSquigglebottom

Right

John Rennie

It's the angle of the velocity to the horizontal as the particle moves along the parabola.

So no θ is not already known.

Binky McSquigglebottom

👍👍👍

John Rennie

And if we now substitute for sinθ we get:
m d²y/dt² = -mg - α√(vx² + vy²) × vy/√(vx² + vy²)
and the square root cancels out.

This leaves:
m d²y/dt² = -mg - αvy

Binky McSquigglebottom

Clear

John Rennie

08:49

And vy is dy/dt, so we end up with:
m d²y/dt² = -mg - α dy/dt

Binky McSquigglebottom

Yes

John Rennie

I won't go through it, but we can do the same with the x direction and we get:
m d²x/dt² = - α dx/dt

Binky McSquigglebottom

Why does m appear on the left?

d²x/dt² = Fx

John Rennie

It's the second law F = ma, but written as ma = F

So on the left side we have m d²y/dt² and on the right side we have F.

Binky McSquigglebottom

@JohnRennie Maybe you missed it here, right?

John Rennie

08:52

Oops, yes, that was a typo.

Binky McSquigglebottom

Was the question for safety

John Rennie

It's probably tidier to divide through by m because we get:
d²y/dt² = -g - α/m dy/dt
d²x/dt² = - α/m dx/dt

Binky McSquigglebottom

So these would be the acceleration formulas?

John Rennie

These are the differential equations we need to solve to get y(t) and x(t)

In fact these can be solved easily.

When I said they were hard to solve I was getting mixed up and thinking of quadratic drag.

Binky McSquigglebottom

But so to find the other expressions, such as the range, the trajectory, do I necessarily have to solve them?

John Rennie

09:01

Yes, because you need to find y(t) so you can calculate how long the particle stays in the air.

Then when you find the time T that the particle falls back to the ground the range is just x(T).

Binky McSquigglebottom

@JohnRennie I don't know how to solve them, I haven't learn them 😔

John Rennie

Differential equations are often solved by guessing what the solution is and then substituting our guess to see what happens. When you learn differential equations at college you'll learn all about the various types of guesses.

Suppose we take the x equation:
d²x/dt² = - α/m dx/dt

Binky McSquigglebottom

Ok

John Rennie

dx/dt is the x component of the velocity vₓ so we can write this as:
dvₓ/dt = -α/m vₓ
Yes?

Binky McSquigglebottom

Yes

Ah but now we do the integral

John Rennie

09:07

Yes, let's rearrange this as:
dvₓ/vₓ = -α/m dt

And then integrate both sides:
∫dvₓ/vₓ = -α/m ∫dt

OK so far?

Binky McSquigglebottom

Yes

John Rennie

On the left side we have vₓ⁻¹ and that integrates to ln(vₓ)

And on the right side we have dt and that integrates to t

Binky McSquigglebottom

Clear

John Rennie

And we need to add a constant of integration, so we end up with:
ln(vₓ) = -α/m t + C

Yes?

Binky McSquigglebottom

Yes

John Rennie

09:11

Taking exp() of both sides we get:
vₓ = A exp(-α/m t)
where A = exp(C)

And to find A we use the initial conditions i.e when t = 0 we have vₓ = vₓ₀
where vₓ₀ is the initial horizontal component of the veocity.

Yes?

Binky McSquigglebottom

Why is there * on the right now?

John Rennie

exp(a + b) = exp(a) × exp(b)
Yes?

Binky McSquigglebottom

Wait

John Rennie

Yes ... ?

Binky McSquigglebottom

I find e^(-a/m • t)+c

I was wrong

@JohnRennie yes

It's like you did

John Rennie

09:16

OK :-)

So now we know the expression for vₓ as a function of time:
vₓ(t) = vₓ₀ exp(-α/m t)

Yes?

Binky McSquigglebottom

Yes

John Rennie

And we can just integrate vₓ(t) to get x(t)

x(t) = -vₓ₀ m/α exp(-α/m t) + C

where C is the constant of integration. And to find C we use the initial condition x = 0 when t = 0.

Yes?

Binky McSquigglebottom

Yes

John Rennie

If we substitute x = 0, t = 0 we get:
0 = -vₓ₀ m/α + C

So C = vₓ₀ m/α

Binky McSquigglebottom

Yes

John Rennie

09:24

So the full equation is:
x(t) = vₓ₀ m/α - vₓ₀ m/α exp(-α/m t)

Or we can tidy this up to get:
x(t) = vₓ₀ m/α (1 - exp(-α/m t))

Binky McSquigglebottom

Yes

John Rennie

So, it was a lot more complicated than the trajectory without drag, but we can solve it and the final equation is not too awful.
Yes?

Well, the final equation for x(t). We still have to find y(t) ...

Binky McSquigglebottom

We can always write it like we did for this one

dvy/dt

John Rennie

d²y/dt² = - g - α/m dy/dt

So we get:
dvy/dt = -g - α/m vy

Binky McSquigglebottom

Yes

John Rennie

09:31

But I'm running out of energy now. Let's look at this tomorrow.

Binky McSquigglebottom

Ok! 👍

Anyway

John Rennie

The problem is that in real life the drag is proportional to v² not v.

Binky McSquigglebottom

Thanks so much for the help

John Rennie

And if we try to solve the equations for this we find they cannot be solved.

@BinkyMcSquigglebottom You're welcome :-)

Pizza

Hi, can I ask a question too? Or better tomorrow?

John Rennie

09:36

Ask now!

@Pizza Are you going to ask?

Pizza

What happend in Konig's theorem if the radius and the velocity are parallel in angular momentum

and what happend at the velocity of the center of mass if the reference system is that of the center of mass

John Rennie

I must admit I don't know König's theorem. I assume you mean this:

König's theorem (kinetics)

In kinetics, König's theorem or König's decomposition is a mathematical relation derived by Johann Samuel König that assists with the calculations of angular momentum and kinetic energy of bodies and systems of particles. The theorem is divided in two parts. The first part expresses the angular momentum of a system as the sum of the angular momentum of the centre of mass and the angular momentum applied to the particles relative to the center of mass. L →...

Pizza

Yes

John Rennie

So when you say if the radius and the velocity are parallel do you mean the radius and velocity of the individual particle relative to the centre of mass?

Pizza

Yes

John Rennie

09:46

The individual angular moment of each particle is:
Lᵢ = rᵢ × mᵢvᵢ
where × is the vector cross product. Yes?

Pizza

Yes

John Rennie

If rᵢ and vᵢ are parallel then the cross product is zero.

So in that case Lᵢ = 0

Pizza

Because theta between the two vectors would be 0

And so sin(0) = 0?

John Rennie

yes

Pizza

11 mins ago, by Pizza

and what happend at the velocity of the center of mass if the reference system is that of the center of mass

So it's 0

John Rennie

09:53

Yes

I need to go. I will be around tomorrow as usual.

Pizza

👍, Thanks so much for the clarification

Binky McSquigglebottom

@JohnRennie 👋

John Rennie

You're welcome. Bye :-)

Transcript for

Problem Solving Strategies