Conductor problem in which the conductor consists of two parallel plates

[Benzoate]

Homework Statement

In a vacuum diode, electrons are "boiled" off a hot cathode, at potential zero, and accelerated across a gap to the anode , which is held at positive potential V0. The cloud of moving electrons within the gap(called space charge) quickly builds up to the point where it reduces the field at the surface of the cathode to zero. From then on a steady current I flows between the plate.

Suppose the plates are large relative to the separation(A>>d^2), so that edge effects can be neglected. Then V, rho, and v(the speed of the electrons) are all functions of x alone.

a) Write Poisson's equation for the region between the plates.

Homework Equations

$$\nabla^2$$ V= $$\nabla$$ $$\cdot$$ E ; If I did not format latex correctly , then here is what I alway trying to write out below:

del^2 V= del E

V=$$\int$$ E $$\cdot$$ dl

The Attempt at a Solution

to get V , I easily integrate E dot dl. How would I obtain the electric field of a parallel plate capacitor. I don't need V to obtain the Poisson equation since:

$$\nabla^2$$ V = $$\nabla$$ $$\cdot$$ E

once I calculate E, how would I find the divergence of E? would I set my coordinate system to a cartesian coordinate?

 

[gabbagabbahey]

I think I recognize this question from Griffith's. Part (a) is supposed to be very simple:

Poisson's equation is always:

$${\nabla}^2V=\frac{\rho}{{\epsilon}_0}$$

But you're told that $$V$$ is a function of $$x$$...so what is $${\nabla}^2V$$?

 

[Benzoate]

I think I recognize this question from Griffith's. Part (a) is supposed to be very simple:

Poisson's equation is always:

$${\nabla}^2V=\frac{\rho}{{\epsilon}_0}$$

But you're told that $$V$$ is a function of $$x$$...so what is $${\nabla}^2V$$?

rho=M/V=M/(x^2*d)

since V=A*d =x^2*d , I am not really sure if the shape of each of the parallel plate is a square , so I can't be sure that A=x^2. If a parallel plate is a square , then rho/epsilon=(M/x^2*d)/epsilon.

 

[gabbagabbahey]

It's actually much simpler than that:

In cartesian coordinates,

$${\nabla}^2V=\frac{d^2V}{dx^2}+\frac{d^2V}{dy^2}+\frac{d^2V}{dz^2}$$

But what are
$$\frac{d^2V}{dy^2}, \quad \frac{d^2V}{dz^2}$$
if V is a function of x alone?

 

[Benzoate]

It's actually much simpler than that:

In cartesian coordinates,

$${\nabla}^2V=\frac{d^2V}{dx^2}+\frac{d^2V}{dy^2}+\frac{d^2V}{dz^2}$$

But what are
$$\frac{d^2V}{dy^2}, \quad \frac{d^2V}{dz^2}$$
if V is a function of x alone?[/


If you are taking the second derivative in terms of x , y and z would be treated as constants. Therefore, $$\frac{d^2V}{dy^2}=0, \quad \frac{d^2V}{dz^2}=0$$

Would I just leave $$\frac{d^2V}{dx^2}$$ as it is ?

 

[gabbagabbahey]

Yes, part (a) just asks you to write poisson's equation , not solve it.

 

[Benzoate]

I have another question to add:

b) Assuming the electrons start from rest at the cathode, what is their speed at point x, where x is V(x)?

W=1/2*CV^2, Work also equals W= 1/2*mv^2

W(capacitor)= W(Kinetic); therefore wouldn't v=sqrt(CV^2/m)

 

[gabbagabbahey]

Yes, but what is the definition of capacitance ,$$C$$ ?

 

[Benzoate]

Yes, but what is the definition of capacitance ,$$C$$ ?

C=Q/V ; That means I calculate my velocity in terms of m , V and Q right?

 

[gabbagabbahey]

Yes, since C is strictly a geometric property of the plates, and you are not given C, it is better to express v in terms of V, m, q; all quantities that you are either given or can calculate.

 

[Benzoate]

Yes, since C is strictly a geometric property of the plates, and you are not given C, it is better to express v in terms of V, m, q; all quantities that you are either given or can calculate.

I will post part c and part d of this question:

part c) In the steady state, I is independent of x. What , then , is the relation between rho an v?

I use equation for Drift velocity :

(n*A*v_p $$\Delta$$ t)= $$\Delta$$Q not sure how to obtain relation between rho and v.

part d) Use these three results(part a, b, and c?) to obtain a differential equation for V by eliminating rho and v.

Not sure where to begin on this part of the problem.

 

[gabbagabbahey]

part c) In the steady state, I is independent of x. What , then , is the relation between rho an v?

I use equation for Drift velocity :

(n*A*v_p $$\Delta$$ t)= $$\Delta$$Q not sure how to obtain relation between rho and v.

You're sort of on the right track here; What is the definition of current (I'm looking for a very simple differential equation here ;))?

 

[Benzoate]

You're sort of on the right track here; What is the definition of current (I'm looking for a very simple differential equation here ;))?

We really haven't studied current yet; but I remember from my intro physics class that I=dq/dt right?

 

[gabbagabbahey]

Right, now what is the relationship between q and rho?

 

[Benzoate]

Right, now what is the relationship between q and rho?

isn't rho the charge density?

 

[gabbagabbahey]

Yes, so what is the mathematical relationship (I'm looking for a simple integral equation)

 

[Benzoate]

Yes, so what is the mathematical relationship (I'm looking for a simple integral equation)

J= $$\sigma$$*v_drift, J being the current density, $$\sigma$$ being density of charge per volume.

 

[gabbagabbahey]

I was simply looking for this relationship:

$$q=\int_{\mathcal{V}}\rho (\vec{r'}) d\tau'$$

And since rho is only a function of x in between the plates during the steady state,

$$q= A \int \rho (x) dx$$

What does this make I=dq/dt?

Hint: use the chain rule :
$$\frac{dq}{dt}=\frac{dq}{dx} \frac{dx}{dt}$$

 

[Benzoate]

I was simply looking for this relationship:

$$q=\int_{\mathcal{V}}\rho (\vec{r'}) d\tau'$$

And since rho is only a function of x in between the plates during the steady state,

$$q= A \int \rho (x) dx$$

What does this make I=dq/dt?

q=A*rho*dx=A*rho*dx

I=$$\delta$$ Q/($$\delta$$ t)

I=A*$$\rho$$*v_drift since, v_drift = dx/dt

 

[gabbagabbahey]

Yes,

$$I=\frac{dq}{dt}=\frac{dq}{dx} \frac{dx}{dt}=A \rho (x) v(x)$$

Now, what do you get for part (d) using this and parts (a) and (b)?

 

[Benzoate]

Yes,

$$I=\frac{dq}{dt}=\frac{dq}{dx} \frac{dx}{dt}=A \rho (x) v(x)$$

Now, what do you get for part (d) using this and parts (a) and (b)?

the problem says that I need to write out a differential equation where $$\rho$$ andv are eliminated

v= $$\sqrt{2*Q*V/m}$$
dq/dt=A*$$\rho$$*v
$$\rho$$=Q/Volume

dq/dt=A**$$\(Q/Volume)$$*$$\sqrt{2*Q*V/m}$$

 

[gabbagabbahey]

That's not really the DE the problem is looking for:

In part (a) you showed
$$\frac{d^2V}{dx^2}=\frac{-\rho}{\epsilon_0}$$

Use the result of part (c) to eliminate $\rho$ in favor of $v$ (Remember I and A are constants so just leave them as is)

Then use (b) to eliminate $v$ in favor of $V$. This way you will have a DE involving only potential and a bunch of constants.

 

[Benzoate]

That's not really the DE the problem is looking for:

In part (a) you showed
$$\frac{d^2V}{dx^2}=\frac{-\rho}{\epsilon_0}$$

Use the result of part (c) to eliminate $\rho$ in favor of $v$ (Remember I and A are constants so just leave them as is)

Then use (b) to eliminate $v$ in favor of $V$. This way you will have a DE involving only potential and a bunch of constants.

I would integrate $$\frac{d^2V}{dx^2}=\frac{-\rho}{\epsilon_0}$$
twice and V=-$$\rho$$*x²/(2*$$\epsilon_0$$==> $$\rho$$=2*V*$$\epsilon_0$$/(x²)

v=$$\sqrt{2*Q*V/(m)}$$

dq/dt=A*$$\rho$$=2*V*$$\epsilon_0$$/(x²)*$$\sqrt{2*Q*V/(m)}$$

 

[gabbagabbahey]

But rho is an unknown function of x, not a constant. You can't treat it as a constant when integrating. Instead you should use the results of (b) and (c) to eliminate it from your ODE.

 

[Benzoate]

should I integrate dq/dt= A*$$\rho$$*v ==>$$\delta$$Q= A*$$\rho$$*v*$$\delta$$t

 

[gabbagabbahey]

No. The question asks you to find a differential equation for V(x) not Q(x,t)! Let's start with step one:

You have: $I=A \rho(x) v(x)$. Solve this for $\rho (x)$

 

[Benzoate]

No. The question asks you to find a differential equation for V(x) not Q(x,t)! Let's start with step one:

You have: $I=A \rho(x) v(x)$. Solve this for $\rho (x)$

$$\rho$$= I/(A*v)

dq/dt= A*$$\rho$$*v=A*(I/(A*v))*v= I

 

[gabbagabbahey]

$$\rho (x)=\frac{I}{Av(x)}$$

so substitute this into your result from (a) and you get:

$$\frac{d^2V}{dx^2}=\frac{-I}{\epsilon_0 Av(x)}$$

Now substitute the result of (b) into this (for v) and what do you get?

 

[Benzoate]

$$\rho (x)=\frac{I}{Av(x)}$$

so substitute this into your result from (a) and you get:

$$\frac{d^2V}{dx^2}=\frac{-I}{\epsilon_0 Av(x)}$$

Now substitute the result of (b) into this (for v) and what do you get?

$$\frac{d^2V}{dx^2}=\frac{-I}{\epsilon_0 Av(x)}$$= $$\frac{d^2V}{dx^2}=\frac{-I}{\epsilon_0 A\sqrt{2*Q*V/m}}$$

 

[gabbagabbahey]

Yes, you now have an ODE for V(x) involving only V(x), its derivatives, and some constants; so there is your answer to part (d). For part (e) you are asked to solve this ODE for V(x), to make it easier to work with I recommend you collect all your constants into one, say,

$$\kappa \equiv \frac{-I}{\epsilon_0 A} \sqrt{\frac{m}{2q}}$$

$$\Rightarrow \frac{d^2V}{dx^2}= \kappa V^{-1/2}$$

Now, try to solve this ODE.