The problem in the image asks to calculate the potential difference (ΔV) between two plates, given the following parameters:

• Area of the plates: $A = 5.00 \times 10^{-3} \, \text{m}^2$
• Separation distance: $d = 4.80 \times 10^{-4} \, \text{m}$
• Charge moved from one plate to the other: $Q = 4.40 \times 10^{-8} \, \text{C}$

The plates form a parallel plate capacitor, and the potential difference between the plates is given by:

$\Delta V = \frac{Qd}{\epsilon_0 A}$

where:

• $\epsilon_0$ is the permittivity of free space: $\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}$.

Now, let's calculate the potential difference using the given values.The potential difference (ΔV) between the two plates is approximately $477.29 \, \text{V}$.

Would you like further details or have any questions?

Here are 5 related questions you may find interesting:

1. How does increasing the separation distance between the plates affect the potential difference?
2. What is the relationship between the electric field and the potential difference in a parallel plate capacitor?
3. How would the potential difference change if the charge were doubled?
4. What role does the permittivity of free space play in the calculation of capacitance?
5. How can the capacitance of the plates be calculated from the given information?

Tip: For parallel plate capacitors, the potential difference is directly proportional to the distance between the plates and the charge on the plates.