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The potential difference \(V(t)\) between parallel plates shown above is instantaneously increasing at a rate of \(10^{7} \mathrm{V} / \mathrm{s} .\) What is the displacement current between the plates if the separation of the plates is \(1.00 \mathrm{cm}\) and they have an area of \(0.200 \mathrm{m}^{2}\) ?

Short Answer

Expert verified
The displacement current will be obtained by inserting the computed rate of change of electric flux and the permittivity constant value into the displacement current formula.

Step by step solution

01

Determine the electric field

The electric field \(E\) between parallel plates can be calculated by dividing the potential difference \(V\) by the separation \(d\) of the plates. Here, \(V\) is changing with time \(t\), so we will have \(E(d,t) = V(t)/d\). Since \(d\) is constant, its rate of change is zero. Given that \(V\) is changing at \(10^{7} \mathrm{V}/\mathrm{s}\), we can write it like this \(E(d,t) = 10^{7} \mathrm{V}/\mathrm{s} / 0.01 \mathrm{m}\)
02

Calculate electric flux

The electric flux \(\Phi_E \) is given by the product of the electric field \(E\) and the area \(A\). So, we have \(\Phi_E = E \cdot A = (10^{7} \mathrm{V}/\mathrm{s} / 0.01 \mathrm{m}) \cdot (0.20 \, \mathrm{m}^2)\). This will give us electric flux as a function of time.
03

Determine the displacement current

Now we can calculate the displacement current \(I_D\) using the formula \(I_D = \varepsilon_0 \cdot \frac{d\Phi_{E}}{dt}\). The permittivity of free space \(\varepsilon_0\) is \(8.85 \times 10^{-12} \, \mathrm{C}^2/\mathrm{N} \cdot \mathrm{m}^2\). Here, \(\frac{d\Phi_{E}}{dt}\) is the rate of change of electric flux which we computed in the previous step. Plugging in the values will yield the displacement current.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Field
Imagine the space between two charged plates; the influence they exert on other charges in the region is known as the electric field. It's a vector field that associates a vector to every point in space, pointing in the direction a positive test charge would move if placed there. In the context of our exercise, the electric field is especially relevant because it underlies the concept of a potential difference, or voltage, causing charges to move from one plate to another.

The strength of an electric field between parallel plates, such as in the given exercise, is uniform and can be calculated easily as the potential difference divided by the distance between the plates. With a changing voltage, the electric field fluctuates over time, influencing how charges behave in that space.
Electric Flux
When discussing electric fields, the concept of electric flux becomes central. It quantifies the number of electric field lines passing through a given area. The electric flux is directly proportional to the strength of the electric field and the size of the area it's passing through. Therefore, a larger area or stronger electric field results in greater electric flux.

In our exercise, this concept is used to describe how the electric field's influence changes over time as the potential difference between the plates changes. By knowing the area of the plates and the time-varying electric field, we can calculate the electric flux at any given moment. This quantity is essential for subsequently determining the displacement current.
Permittivity of Free Space
To fully understand how electric fields interact with the vacuum or air between the plates, we need to introduce the permittivity of free space, denoted as \( \varepsilon_0 \). This constant represents how much electric field (force) is 'permitted' to pass through free space. It’s one of nature's fundamental constants and plays a key role in the equations of electromagnetism.

In practical terms, \( \varepsilon_0 \) affects the storage of electric energy within an electric field and relates to phenomena such as capacitance and displacement current, which we encounter in the given problem. It sets the scale for the displacement current calculation, thereby linking the changing electric flux through the area between the plates to the displacement current.
Potential Difference
Lastly, we come across the term potential difference, also known as voltage. It's a measure of the work done by an electric field to move a charge from one point to another. In simple words, potential difference is the 'push' that compels charges to move across a circuit or, as in our case, between two plates.

In the described exercise, this 'push' is increasing over time. The rate of this increase (\(10^7 \mathrm{V}/\mathrm{s}\)) directly impacts the electric field between the plates. By understanding potential difference and its rate of change, we construct the bridge between the abstract concept of an electric field and the physical flow of electricity – culminating in the notion of displacement current between the plates.

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