To find the potential difference ($V$) between the plates of a parallel-plate capacitor, we use the formula:

$V = \frac{Q}{C}$

where:

• $Q$ is the charge on the plates, and
• $C$ is the capacitance of the capacitor.

Step 1: Capacitance of the Capacitor

The capacitance $C$ of a parallel-plate capacitor with a dielectric is given by:

$C = \frac{\kappa \varepsilon_0 A}{d}$

where:

• $\kappa = 2.1$ is the dielectric constant,
• $\varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m}$ is the permittivity of free space,
• $A = 1.3 \times 10^{-2} \, \text{m}^2$ is the area of the plates,
• $d = 0.80 \, \text{mm} = 0.80 \times 10^{-3} \, \text{m}$ is the separation between the plates.

Let's calculate $C$ first.

$C = \frac{(2.1)(8.854 \times 10^{-12})(1.3 \times 10^{-2})}{0.80 \times 10^{-3}}$

Step 2: Calculate Capacitance

Now, we compute the value of $C$.The capacitance $C$ of the capacitor is approximately $3.02 \times 10^{-10} \, \text{F}$.

Step 3: Calculate the Potential Difference

Now that we have the capacitance, we can find the potential difference using the formula:

$V = \frac{Q}{C}$

where $Q = 4.1 \, \mu C = 4.1 \times 10^{-6} \, C$.

Substituting the values:

$V = \frac{4.1 \times 10^{-6}}{3.02 \times 10^{-10}}$

Let's calculate $V$.The potential difference between the plates is approximately $13,570 \, \text{V}$.

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. How does changing the dielectric constant affect the potential difference?
2. What happens to the capacitance if the separation between the plates is halved?
3. How would the potential difference change if the area of the plates is doubled?
4. What is the energy stored in the capacitor with this charge and potential difference?
5. How does the material of the dielectric impact the electric field between the plates?

Tip: The capacitance of a parallel-plate capacitor increases if a dielectric with a higher constant is used.