# Not understanding work stored in a capacitor

• yosimba2000
In summary, the work done by the demon in moving charge from one plate to the other is equal to the integral of q/c dq.
yosimba2000
I know V = q/c and W = Vq and dW = V dq. But why is Work in charging a capacitor W = integral of q/c dq?

q seems to represent a charge on the capacitor plate and dq seems to represent a separate test charge. If I add a charge to the capacitor plate, I take take the resulting votage and multiply by test charge dq to get the work done on dq. For each additional charge added to the capacitor, I continue to multiply the new voltage by dq. But why should I multiply by dq? The voltage created affects any amount of charge, so why is it restricted to affecting only dq?

For the basic Energy equation E = integral of F(x)dx, this means I am splitting up distance x into pieces of size dx, calculating the force at each segment, multiplying it by the distance dx over which the force acts, and adding everything up. I can see why F(x) is related to dx because each value of F(x) only works across distance dx.

The work charging a capacitor is similar at least mathematically to the work done in compressing a spring and therefore storing energy in the spring.

for a spring dw = k⋅x⋅dx where x is the present amount of compression and k is the spring constant basically how much the spring resists compression per unit of compression.

So W = k ∫xdx = ½kx2 just as work stored in a capacitor as energy is ½Cq2

Dullard and berkeman
Yes, that's where I am getting caught up. For a spring, each small distance you move affects the spring force, and that force is applied specifically for that small distance dx.

But for a capacitor, each small charge you add to the plate increases the voltage, but why is that voltage applied for only a small test charge dq? Why would that test charge dq affect the value of the next charge that arrives at the capacitor plate?

yosimba2000 said:
But for a capacitor, each small charge you add to the plate increases the voltage, but why is that voltage applied for only a small test charge dq? Why would that test charge dq affect the value of the next charge that arrives at the capacitor plate?

In the spring each increment in the compression increases the resisting force that occurs for that particular small increment. Each succeeding increment results in a stronger resisting force. Similarly in the capacitor each incremental increase in charge for a capacitor increases the voltage and therefore the electric field that makes it harder for the next incremental increase in charge to be placed in the capacitor.

yosimba2000 said:
Ok, so dq represents the incremental charge to be applied on the capacitor plate and not an imaginary test charge between the plates?

Yes. I do not know where you got the imaginary test charge from or how it is used to describe this process.

The general idea is the following.

Suppose a usual parallel-plate capacitor and as a gedanken experiment consider a demon, shuffeling (positive) infinitesimal charges from one plate to the other, starting from both plates being neutral up to the final state, where there is a total charge ##Q## at one plate and ##-Q## at the other. Suppose that the demon works very slowly, so that you can neglect the acceleration of the charges, i.e., we neglect the radiation of accelerated charges.

Suppose the demon is at the point, where there is an amount of charge ##q## on one of the plates and ##-q## at the other. Then carrying the next portion of charge, ##\mathrm{d} q##, to the other plate he has to do work against the electric field.With the capacitance ##C## the electric potential-difference between the plates is ##U=q/C## and thus the total work he has to do is ##\mathrm{d} W=\mathrm{d}q \, q/C##. Now you have to "sum" over this entire process. In the limit of making ##\mathrm{d} q## infinitesimally small the sum becomes
$$W=\int_0^{Q} \mathrm{d} q \frac{q}{C} = \frac{Q^2}{2C}=\frac{C}{2} U^2.$$

## 1. What is a capacitor?

A capacitor is an electronic component that stores electric charge. It consists of two conductive plates separated by an insulating material, called a dielectric.

## 2. How does a capacitor store energy?

When a voltage is applied across a capacitor, it creates an electric field between the plates. This field causes electrons to accumulate on one plate and be removed from the other plate, creating a potential difference. The energy is stored in this electric field.

## 3. Why is it important to understand how a capacitor works?

Capacitors are used in a wide range of electronic devices and circuits, so understanding how they work is crucial for designing and troubleshooting these systems. Additionally, capacitors have unique properties that make them useful in specific applications, so understanding their operation allows for the selection of the right capacitor for a given task.

## 4. How do I calculate the amount of energy stored in a capacitor?

The amount of energy stored in a capacitor can be calculated using the formula E = 1/2 * C * V^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage across the capacitor in volts.

## 5. What factors affect the amount of energy stored in a capacitor?

The amount of energy stored in a capacitor is affected by its capacitance, the voltage applied, and the type of dielectric material used. The distance between the plates, the surface area of the plates, and the temperature also play a role in the amount of energy stored. Additionally, the type of capacitor (e.g. electrolytic, ceramic, etc.) can affect the amount of energy it can store.

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