The electric field between the two plates of a parallel plate (area = 5.0 × 10-2m2) capacitor is given by E = (2.0 × 105 – 4.0 × 104 t) V/m where t is in second. What is the magnitude of the displacement current between the plates? (ε0 = 8.85 × 10-12 F/m)
1.77 × 10-8 A
Displacement Current in a Parallel Plate Capacitor
The problem requires us to calculate the magnitude of the displacement current between the plates of a parallel plate capacitor, given its area and the time-varying electric field between its plates. This involves applying a fundamental concept from electromagnetism related to Maxwell's equations.
Understanding Displacement Current
The displacement current (\(I_D\)) is a concept introduced by James Clerk Maxwell. Unlike conduction current, which is due to the flow of charges, displacement current arises from a changing electric field or, more specifically, a changing electric flux. It plays a crucial role in Maxwell's equations, particularly in modifying Ampere's circuital law to make it consistent for time-varying fields and to explain the propagation of electromagnetic waves.
- Electric Flux (\(\Phi_E\)): This is a measure of the total electric field passing through a given area. For a uniform electric field \(E\) passing perpendicularly through a flat surface of area \(A\), the electric flux is simply the product of the electric field strength and the area: \(\Phi_E = E \cdot A\).
- Permittivity of Free Space (\(\epsilon_0\)): This physical constant describes the ability of a vacuum to permit electric field lines. Its value is approximately \(8.85 \times 10^{-12} \text{ F/m}\).
Formula for Displacement Current
The formula for displacement current is derived from Maxwell's extension of Ampere's circuital law. It states that the displacement current \(I_D\) is proportional to the rate of change of electric flux (\(\frac{d\Phi_E}{dt}\)) through a surface:
\[ I_D = \epsilon_0 \frac{d\Phi_E}{dt} \]
For a parallel plate capacitor, where the electric field \(E\) is uniform between the plates and perpendicular to the plate area \(A\), the electric flux \(\Phi_E\) is \(E \cdot A\). Since the area \(A\) of the plates remains constant, we can substitute this into the formula:
\[ I_D = \epsilon_0 \frac{d(E \cdot A)}{dt} \]
\[ I_D = \epsilon_0 A \frac{dE}{dt} \]
This modified formula allows us to calculate the displacement current when the electric field between the capacitor plates changes over time.
Given Parameters
Let's list the values provided in the question that are necessary for our calculation:
| Parameter | Symbol | Value |
|---|---|---|
| Area of the parallel plate capacitor | \(A\) | \(5.0 \times 10^{-2} \text{ m}^2\) |
| Electric field between the plates | \(E\) | \((2.0 \times 10^5 - 4.0 \times 10^4 t) \text{ V/m}\) |
| Permittivity of free space | \(\epsilon_0\) | \(8.85 \times 10^{-12} \text{ F/m}\) |
Step-by-Step Displacement Current Calculation
To find the magnitude of the displacement current, we follow these steps:
- Calculate the rate of change of the electric field (\(\frac{dE}{dt}\)):
The electric field \(E\) is given as a function of time \(t\):
\[ E = (2.0 \times 10^5 - 4.0 \times 10^4 t) \text{ V/m} \]
To find its rate of change, we differentiate \(E\) with respect to time \(t\):
\[ \frac{dE}{dt} = \frac{d}{dt} (2.0 \times 10^5 - 4.0 \times 10^4 t) \]
The derivative of a constant (\(2.0 \times 10^5\)) is zero. The derivative of \(-4.0 \times 10^4 t\) with respect to \(t\) is simply \(-4.0 \times 10^4\).
\[ \frac{dE}{dt} = -4.0 \times 10^4 \text{ V/m s} \]
- Substitute the values into the displacement current formula:
Now, we use the formula \(I_D = \epsilon_0 A \frac{dE}{dt}\) and plug in the values for \(\epsilon_0\), \(A\), and \(\frac{dE}{dt}\):
\[ I_D = (8.85 \times 10^{-12} \text{ F/m}) \times (5.0 \times 10^{-2} \text{ m}^2) \times (-4.0 \times 10^4 \text{ V/m s}) \]
Multiply the numerical parts and the powers of 10 separately:
\[ I_D = (8.85 \times 5.0 \times -4.0) \times (10^{-12} \times 10^{-2} \times 10^4) \text{ A} \]
\[ I_D = (-177) \times (10^{-12 - 2 + 4}) \text{ A} \]
\[ I_D = -177 \times 10^{-10} \text{ A} \]
To express this in standard scientific notation, we adjust the decimal point:
\[ I_D = -1.77 \times 10^{-8} \text{ A} \]
- Determine the magnitude of the displacement current:
The question specifically asks for the magnitude of the displacement current. The magnitude is the absolute value of the current, which means we disregard the negative sign (which indicates direction or phase).
\[ \text{Magnitude of } I_D = | -1.77 \times 10^{-8} \text{ A} | \]
\[ \text{Magnitude of } I_D = 1.77 \times 10^{-8} \text{ A} \]
Final Result
The magnitude of the displacement current between the plates of the parallel plate capacitor is \(1.77 \times 10^{-8} \text{ A}\). This calculation demonstrates the application of fundamental electromagnetic principles to determine currents induced by changing electric fields.
Important Questions from Electric Fields and Gauss' Law
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If a free electron moves through a potential difference of 1 kV, then the energy gained by the electron is given by
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