Total energy of charge in parallel plate capacitor

[Inquisitive Student]

Homework Statement

Find the total amount of energy of a charge (q) initially at rest placed at the left plate of a parallel plate capacitor. The left plate is at V = 0, the right plate at V = V₀ and the plates are a distance L apart.

Homework Equations

E_tot = 1/2 mv² + qφ(x).
φ(x) = Vx/L

The Attempt at a Solution

Since energy is conserved:

E = E_initial = 1/2 m(v_initial)² + qφ(0) = 0 + 0 = 0.

(I chose x = 0 to be the left plate)
So the total energy that the charge has is 0 everywhere? How does the charge even move at all if it has no energy to begin with?

 

[NFuller]

So the total energy that the charge has is 0 everywhere?

Remember that the electric potential is invariant under addition of a constant. All that matters is how the potential changes with distance. Since $E_{tot}$ is related to how the potential is defined, the total energy is an arbitrary constant. Here, you are using the convention that the potential is zero at the surface of the left plate. Since the kinetic energy at this location is also zero, the total energy is a constant zero. The kinetic energy however, is not constant as the charge moves.

How does the charge even move at all if it has no energy to begin with?

The charge will move in the direction of decreasing potential. If $q$ is negative, this means the charge will move towards the right plate. As this happens, the kinetic energy must increase to preserve the total energy and keep $E_{tot}=0$.

 

[haruspex]

φ(x) = Vx/L

Please explain this equation and how you are using it (if you are). If you are not, how do you get that Φ(x)=0?

 

[Inquisitive Student]

Please explain this equation and how you are using it (if you are). If you are not, how do you get that Φ(x)=0?

From boundary conditions we know that Φ(x = 0) = 0 (stated in the problem that left wall of parallel plate capacitor is at Φ = 0) and that Φ(x = L) = V. Poisson's equation in space (we neglect the charge in the problem, we are just finding the potential due to the walls) is ∇²Φ = 0, or in 1D: $$\frac{\partial^2 Φ}{ \partial x^2} = 0$$. Since the second derivative is equal to 0, that means the first derivative was equal to a constant. So, $$\frac{\partial Φ}{\partial x} = c$$, where c is a constant. We can solve this differential equation by integrating, $$\frac{\partial Φ}{\partial x} = c$$ becomes ∫∂Φ = ∫c ∂x, which means $$Φ(x) = cx + c_1$$ where c₁ is a constant that came from integration. Now plugging in the boundary conditions:

Φ(0) = 0 gives c₁ = 0
Φ(L) = V gives v = c*L or c = v/L

So in total we get: Φ(x) = cx = vx/L

 

[haruspex]

and that Φ(x = L) = V.

If you mean Φ(L)=V₀, yes, Φ(x)=V₀x/L gives the potential as a function of distance from the left plate, valid for all points between the plates. I was thrown by the reference to V instead of V₀.

So does NFuller's response answer your question?

 

[Inquisitive Student]

If you mean Φ(L)=V₀, yes, Φ(x)=V₀x/L gives the potential as a function of distance from the left plate, valid for all points between the plates. I was thrown by the reference to V instead of V₀.

So does NFuller's response answer your question?

Yeah, NFuller had a very clear response. It was really helpful.