How Is Linear Charge Distribution a Realistic Model in Capacitor Charging?

In summary, the problem is that the charge distribution for a capacitor is a linear function of time, which is not realistic.
  • #1
E_M_C
43
0
The problem is stated:

The preceding problem was an artificial model for the charging capacitor, designed to avoid complications associated with the current spreading out over the surface of the plates. For a more realistic model, imagine thin wires that connect to the centers of the plates (the plates are circular). Again, the current I is constant, the radius of the capacitor is a, and the separation of the plates is w « a. Assume that the current flows out over the plates in such a way that the surface charge is uniform, at any given time, and is zero at t = 0.

Find the electric field between the plates, as a function of t.


I understand that I have to use Gauss' Law to find the E-field, but first I have to find the charge distribution Q(t), this is where I'm having some difficulty. After a lot of frustration, I peeked at part of the solution, and I found that the charge distribution is Q(t) = It. I assume it was arrived at as follows:

[itex] I = \frac{dQ(t)}{dt} → \int dQ = \int I dt → Q(t) = It[/itex]

I was surprised to see that the charge distribution is a linear function, as I was expecting an exponential expression. Maybe I'm not getting the concept of a "uniform surface charge", but I don't see how a linear charge distribution is a "realistic model." For example, what happens as t → ∞? The charge distribution blows up. How is that realistic?

Any help is appreciated.
 
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  • #2
The charge of the capacitor is the time integral of the current. If the current is constant in time the charge is a linear function of time.
Charge distribution refers to the distribution of charge along the surface of the capacitor plate. It is assumed to be homogeneous: The surface charge density is σ=Q/A where A is the area of a plate, A=a22π. The problem asks the electric field as function of time. Apply Gauss' Law.
Of course, the constant current can not be maintained for infinity: but it is possible for some finite time. There are devices "current generators" which provide constant current.

ehild
 
  • #3
Thanks ehild.

I suppose that I was just over-thinking the problem. The idea of the capacitor being attached to a current source did cross my mind, but I couldn't think past the limitations of a linear solution.

After some more careful thought, I realized that the exponential solution I was expecting would only arise when the capacitor is in series with a resistor.
 
  • #4
E_M_C said:
After some more careful thought, I realized that the exponential solution I was expecting would only arise when the capacitor is in series with a resistor.

You thought it correctly in case you charge the capacitor with a voltage source with constant emf through a resistor. The current will decrease exponentially with time constant RC.

ehild
 
  • #5


I can understand your confusion and frustration with the linear charge distribution in this problem. However, it is important to remember that this is just a simplified model for the charging capacitor and is not meant to be a completely accurate representation of real-world situations.

In this model, the assumption is made that the current flows out uniformly over the surface of the plates, resulting in a uniform surface charge. This means that at any given time, the amount of charge on each plate is directly proportional to the time that has passed since the charging began. This is why the charge distribution is a linear function of time, with the slope being equal to the constant current I.

As you correctly stated, this linear charge distribution is not realistic in the long term, as it would eventually lead to an infinite charge on the plates. However, in this simplified model, we are only concerned with the initial charging process and not the long-term behavior.

In real-world situations, the charge distribution would not be perfectly uniform due to various factors such as resistance in the wires and non-uniformities in the plates. These would result in a more complex charge distribution, which may indeed follow an exponential function. But for the purposes of this problem, the linear charge distribution serves as a good approximation.

In summary, while the linear charge distribution may not be completely realistic, it is a useful model for understanding the initial charging process of a capacitor. It is important to keep in mind that models are simplifications of real-world phenomena and are not meant to be exact representations.
 

What is a linear charge distribution?

A linear charge distribution is a concept in electromagnetism that describes the distribution of electric charge along a straight line. It is often used to model the behavior of charged particles in a certain direction, such as in an electric wire or a charged rod.

How is linear charge distribution different from point charge?

While a point charge is a single, localized charge, a linear charge distribution is spread out along a line. This means that the electric field produced by a linear charge distribution varies along the line, whereas the electric field produced by a point charge is constant at all points in space.

What is the mathematical expression for the electric field of a linear charge distribution?

The electric field E at a point P located at a distance r from a linear charge distribution with charge density λ is given by the equation E = (1/4πε0) * (λ/r), where ε0 is the permittivity of free space. This equation is known as Gauss's law.

Can linear charge distribution be used to calculate the electric field of a three-dimensional object?

No, a linear charge distribution is only applicable for objects with charge distributed along a straight line. To calculate the electric field of a three-dimensional object, other concepts such as surface charge distribution or volume charge distribution must be used.

How is the electric potential related to linear charge distribution?

The electric potential V at a point P located at a distance r from a linear charge distribution with charge density λ is given by the equation V = (1/4πε0) * (λ/r). This means that the electric potential decreases with distance from the linear charge distribution, similar to the electric field.

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