Solve Eletrostatic Problem: Sphere Charge Q0

[Anonymous119]

[Note from mentor -- this post was moved here from another forum]

Hi,

I ran to this problem today and i try to solve it but i can't complete it so if anyone can help me :)

Sphere has a charge Q0. Inside the sphere is vacuum and outside the air. Consume that air has small non zero conductivity s (the inverse of resistivity=1/ρ=s). Find how its charge
will evolve with time Q = Q(T).

Here is my work:

$Q(T) = Q_0-dQ(T)$

$dQ=\int_0^T I(t)\, dt$
where is I=U/R
R=1/s *l/S where l= dr(distance betwen our sphere(r) and sphere (r+dr)) where dr is very small and S=4pi*r^2
U=kQ(t)*(1/r-1/(r+dr))

When i arranged this i got:

I(t)=Q(t)*s/e0 so

$dQ=\frac{s}{e_0} \cdot \int_0^TQ(t)dt$

$Q(T)=Q_0-\frac{s}{e_0}\cdot\int_0^TQ(t)dt$
Where e0 is eletric constant.

And now i don't know how to solve this equation.
So please if anyone know how or see any mistake in my work post here.

Thanks !

 

[ehild]

Hi,

I ran to this problem today and i try to solve it but i can't complete it so if anyone can help me :)

Sphere has a charge Q0. Inside the sphere is vacuum and outside the air. Consume that air has small non zero conductivity s (the inverse of resistivity=1/ρ=s). Find how its charge
will evolve with time Q = Q(T).

Here is my work:

$Q(T) = Q_0-dQ(T)$

$dQ=\int_0^T I(t)\, dt$

dQ means infinitesimal change of the charge, dQ=(dQ/dt) dt= Idt. Do not use d for finite changes. The change of charge can be written as ΔQ.
And you do not need ΔQ. You need the charge as function of time, Q(t).

where is I=U/R
R=1/s *l/S where l= dr(distance betwen our sphere(r) and sphere (r+dr)) where dr is very small and S=4pi*r^2
U=kQ(t)*(1/r-1/(r+dr))

When i arranged this i got:

I(t)=Q(t)*s/e0

Correct so far.

so

[STRIKE]$dQ=\frac{s}{e_0} \cdot \int_0^TQ(t)dt [/STRIKE]$

The current is equal to the charge leaving the sphere in unit time. I=-dQ/dt.

So $dQ/dt= - \frac{s}{ε_0}Q(t)$

That is a separable differential equation for Q(t). You can write it
$$\int{\frac {dQ}{Q}}=- \frac{s}{ε_0}\int{dt} $$
Perform the integration and fit to the initial condition Q=Q₀ at t=0. If you want definite integral from t=0 to t=T you an write $$\int_{Q_0}^{Q(T)}{\frac {dQ}{Q}}=- \frac{s}{ε_0}\int_0^T{dt} $$

ehild

 

[Delta2]

Hm to solve this problem we just use ohm's law and the definition of current? Shouldnt somewhere use maxwell's equations? Wouldnt the charges flowing from the sphere to the outer imaginary spheres alter the electric field because the overall charge density changes in the volume of air around the sphere and not only in the surface of the sphere... But let me guess, we just neglect the charge density of the surrounding volume of air (as well as considering quasi static approach neglecting the terms dB/dt and dE/dt) in order to simplify the problem.

 

[ehild]

Hm to solve this problem we just use ohm's law and the definition of current? Shouldnt somewhere use maxwell's equations? Wouldnt the charges flowing from the sphere to the outer imaginary spheres alter the electric field because the overall charge density changes in the volume of air around the sphere and not only in the surface of the sphere... But let me guess, we just neglect the charge density of the surrounding volume of air (as well as considering quasi static approach neglecting the terms dB/dt and dE/dt) in order to simplify the problem.

It would be better to use the relation between current density and electric field $\vec j = σ \vec E$ (The OP uses the notation s for σ, the conductivity. )

Integrating around the surface of the sphere, we get the current $I =\int{\vec j d \vec A} = σ\int {\vec E d\vec A }= σ Q/ε_0$. The current is the charge flowing through a surface in unit time - the surface is a sphere concentric with the original one: I=-dQ/dt.

ehild

 

[Anonymous119]

dQ means infinitesimal change of the charge, dQ=(dQ/dt) dt= Idt. Do not use d for finite changes. The change of charge can be written as ΔQ.
And you do not need ΔQ. You need the charge as function of time, Q(t).

Correct so far. The current is equal to the charge leaving the sphere in unit time. I=-dQ/dt.

So $dQ/dt= - \frac{s}{ε_0}Q(t)$

That is a separable differential equation for Q(t). You can write it
$$\int{\frac {dQ}{Q}}=- \frac{s}{ε_0}\int{dt} $$
Perform the integration and fit to the initial condition Q=Q₀ at t=0. If you want definite integral from t=0 to t=T you an write $$\int_{Q_0}^{Q(T)}{\frac {dQ}{Q}}=- \frac{s}{ε_0}\int_0^T{dt} $$

ehild

I think you are absolutely right .
I got confused about the marks how dumb.

THANKS !

I got this for the final solution:

$$Q(t)=Q_0\cdot e^{-t\frac{s}{e_0}}$$

Hm to solve this problem we just use ohm's law and the definition of current? Shouldnt somewhere use maxwell's equations? Wouldnt the charges flowing from the sphere to the outer imaginary spheres alter the electric field because the overall charge density changes in the volume of air around the sphere and not only in the surface of the sphere... But let me guess, we just neglect the charge density of the surrounding volume of air (as well as considering quasi static approach neglecting the terms dB/dt and dE/dt) in order to simplify the problem.

I don't think so because it's problem for high school students and they usually don't study Maxwell's equations. I've just finished the first grade of high school so i don't know enough math to unerstand Maxwell's equations completely.

 

[ehild]

I got this for the final solution:

$$Q(t)=Q_0\cdot e^{-t\frac{s}{e_0}}$$

Excellent!

ehild

 

[Delta2]

It would be better to use the relation between current density and electric field $\vec j = σ \vec E$ (The OP uses the notation s for σ, the conductivity. )

Integrating around the surface of the sphere, we get the current $I =\int{\vec j d \vec A} = σ\int {\vec E d\vec A }= σ Q/ε_0$. The current is the charge flowing through a surface in unit time - the surface is a sphere concentric with the original one: I=-dQ/dt.

ehild

Ok i see so we can say that we use Gauss's Law as well.

But, suppose we want to calculate the charge density in the surrounding volume after time t, how we would proceed?