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The electric field in a capacitor plays an essential role. It's a measure of the force that a charge experiences inside the field created by the charged plates. To find the electric field, we use the relationship between charge, the permittivity of space, and the area of the plates. In the given exercise, the electric field \( E \) is calculated with the formula:

• \( E = \frac{Q}{\varepsilon_0 A} \)

where \( Q \) is the charge on the plates, \( \varepsilon_0 \) represents the permittivity of free space (\( 8.854 \times 10^{-12} \) F/m), and \( A \) is the area of the capacitor plates.Inserting the values given:

• \( Q = 9.00 \times 10^{-10} \text{ C} \)
• \( A = 5.00 \times 10^{-4} \text{ m}^2 \)

provides \( E \approx 203.1 \text{ N/C} \). This shows how the strength of the electric field is linked to how much charge is stored and the geometry of the capacitor.

The potential difference across a capacitor is crucial as it essentially tells us how much energy is required to move a unit charge from one plate to the other. This potential difference or voltage \( V \) is calculated using the electric field and the distance between the plates. The relationship is given by:

• \( V = E \times d \)

where \( d \) is the separation between the plates. For the exercise, \( d \) is 2.00 mm or 2.00 \( \times 10^{-3} \) m.Using the previously calculated electric field \( E = 203.1 \text{ N/C} \), we find:

• \( V = 203.1 \times 2.00 \times 10^{-3} = 0.406 \text{ V} \)

This simple relation illustrates how the voltage across a capacitor increases with a stronger electric field or a larger separation between its plates.

Displacement current is a unique concept important in the realm of electromagnetic theory, especially when understanding capacitors in a circuit. It is a "fictitious" current that explains how changing electric fields can give rise to magnetic fields, even where there is no actual charge flow.The displacement current density \( j_D \) is calculated from the rate of change of the electric field:

• \( j_D = \varepsilon_0 \frac{dE}{dt} \)

In this scenario, since both the charging current \( i_C = 1.80 \times 10^{-3} \text{ A} \) and the area \( A = 5.00 \times 10^{-4} \text{ m}^2 \) are constants, \( \frac{dE}{dt} = \frac{i_C}{\varepsilon_0 A} \) stays constant:

• \( \frac{dE}{dt} \approx 4.07 \times 10^{7} \text{ N/C/s} \)

Thus, \( j_D \approx 0.360 \text{ A/m}^2 \).Finally, the total displacement current \( i_D \) becomes:

• \( i_D = j_D \times A \)
• \( i_D = 1.80 \times 10^{-4} \text{ A} \)

This tells us the displacement current equals the conduction current, explaining why both are crucial when analyzing time-varying fields within a capacitor.