# Generating current through increasing capacitance

• Dilemma
In summary, the conversation discusses finding the current I as a dielectric slab is inserted at a constant rate, using variables such as V, K, r, d, Δt, and ε0. The attempted solution includes the equation Q = CV and the correct answer is found to be ∈(r^2)(K-1)V/(dΔt). The mistake in the attempted solution is the missing dx multiplier, and it is also noted that the equation I = dQ/Δt should be corrected to account for the finite amount of time required to insert the slab.
Dilemma
Hello,

1. Homework Statement

Assuming that the dielectric is inserted at a constant rate, find the current I as the slab is inserted.
Express your answer in terms of any or all of the given variables V, K, r, d, Δt, and ε0, the permittivity of free space.

[/B]
C = ε0 * A / d

## The Attempt at a Solution

[/B]
Q = CV
dQ = dC ⋅ V

I = dQ/Δt

C = ∈ ⋅ A / d = ∈⋅(xr)K / d + ∈⋅(r-x)r / d
dC = ∈⋅(r)K / d - ∈⋅r / d = ∈r(K-1)/d
dQ = dC⋅V

However, the correct answer is ∈(r^2)(K-1)V/(dΔt). That means the multiplier "r" is missing. What is wrong with my solution.

Dilemma said:
dC = ∈⋅(r)K / d - ∈⋅r / d = ∈r(K-1)/d
The left side is a differential quantity. So, the right side should also be a differential quantity.

Oops, I have forgotten to include the dx multiplier. Thank you.

Dilemma said:
I = dQ/Δt

dQ is a differential quantity, but Δt is the finite amount of time required to completely insert the slab.
Did you mean to write I = dQ/dt, or maybe I = ΔQ/Δt?

Dilemma

## What is capacitance?

Capacitance is the ability of a component or system to store an electrical charge. It is measured in farads (F) and is determined by the physical characteristics of the component, such as its size, shape, and the material it is made of.

## How is current generated through increasing capacitance?

Current can be generated through increasing capacitance by connecting a power source, such as a battery, to the capacitor. The capacitor will charge up to the same potential as the power source, creating a difference in charge between its two plates. This difference in charge creates an electric field, which can then produce a flow of current when a circuit is completed.

## What are some applications of increasing capacitance to generate current?

Increasing capacitance to generate current is commonly used in electronic devices such as radios, televisions, and computers. It is also used in power grids to regulate voltage levels and in energy storage systems, such as batteries and capacitors, to store and release energy.

## What factors affect the amount of current generated through increasing capacitance?

The amount of current generated through increasing capacitance is affected by several factors, including the voltage of the power source, the capacitance value, the resistance of the circuit, and the frequency of the power source. Higher voltage and capacitance values will generally result in a larger current, while higher resistance and frequency will decrease the current.

## Are there any limitations or downsides to increasing capacitance to generate current?

While increasing capacitance can be a useful way to generate current, there are some limitations and downsides to consider. Capacitors can only store a limited amount of charge, so they are not suitable for long-term or high-power applications. Additionally, they can be expensive and bulky compared to other methods of generating current.

Replies
16
Views
400
Replies
7
Views
1K
Replies
3
Views
2K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
4
Views
4K
Replies
2
Views
1K
Replies
10
Views
3K
Replies
6
Views
2K
Replies
12
Views
812