91Ó°ÊÓ

When discussing capacitors, the electric field is a crucial concept. It describes how the electric force is distributed between the plates of a capacitor. For a parallel-plate capacitor, the electric field (\[ E \] ) is uniform between the plates and can be calculated by the formula:

• \[ E = \frac{V}{d} \]

This formula shows that the electric field strength depends directly on the voltage (\[ V \] ) across the plates and inversely on the distance (\[ d \] ) between them.
This means that if you increase the voltage while keeping the distance constant, the electric field becomes stronger, and vice versa. For Capacitor A, which is filled with air, the electric field is calculated using this formula to be 200,000 V/m.
In Capacitor B, which contains a dielectric, the calculation slightly changes. A dielectric reduces the electric field, which we will explore in the next section.

The introduction of a dielectric material into a capacitor affects its properties significantly. The dielectric constant (\[ \kappa \] ) is a dimensionless quantity that measures a material's ability to reduce the electric field compared to the field in a vacuum.
It does so by aligning its molecular structure in a way that opposes the external electric field. This opposing field effectively decreases the overall electric field within the material.

• The formula for the electric field within a capacitor containing a dielectric is given by:\[ E = \frac{E_{\text{air}}}{\kappa} \]

In our exercise, Capacitor B has a dielectric constant of 2.60, reducing the electric field from 200,000 V/m to approximately 76,923 V/m.
This reduced electric field has implications for the charge capacity and stored energy of the capacitor.

Charge density (\[ \sigma \] ) represents the amount of electric charge per unit area on the surface of the capacitor plates. It plays a vital role in understanding how much charge is being stored in the capacitor for a given electric field.
For Capacitor A, the charge density can be calculated using the formula:

• \[ \sigma_A = \varepsilon_0 \times E \]

Here, (\[ \varepsilon_0 \] ) is the permittivity of free space, approximately equal to \( 8.85 \times 10^{-12} \text{ C/V·m} \).
Using this, we found that the charge density for Capacitor A is \( 1.77 \times 10^{-6} \text{ C/m}^2 \). For Capacitor B, filled with a dielectric, the charge density is calculated at \( 6.8 \times 10^{-7} \text{ C/m}^2 \), reflecting the influence of the dielectric on the charge distribution.

Introducing a dielectric not only reduces the electric field but also affects other properties of the capacitor. The presence of a dielectric increases the capacitor's ability to store charge for the same applied voltage. This is because the dielectric reduces the potential difference for a given charge density, effectively increasing the capacitance.
The induced charge (\[ \sigma' \] ) on the surface of the dielectric itself is a direct result of the dielectric's influence. It can be calculated using:

• \[ \sigma' = \left(1 - \frac{1}{\kappa} \right) \times \sigma_B \]

This formula accounts for the charge that appears on the dielectric, opposing the field created by the free charge density. In our problem, the induced charge density is found to be \( 4.17 \times 10^{-7} \text{ C/m}^2 \), which illustrates the balancing act performed by the dielectric.
This induced charge doesn't add to the net charge on the capacitor but impacts how the capacitor interacts with its surroundings, particularly in energy storage and discharge.