A parallel-plate capacitor is designed so that the plates can be pulled apart. The capacitor is initially charged to a potential difference of 80 Volts when the plates are 1.0 mm apart. The plates are insulated so that the charge cannot leak off. What is the potential difference between the plates when they are pulled to a new separation 3.5 mm apart?

To solve this problem, we need to understand the relationship between charge, potential difference, and capacitance in a parallel-plate capacitor.

The formula for capacitance (C) in a parallel-plate capacitor is given by:

C = ε₀ * (A / d)

Where:
- C is the capacitance
- ε₀ is the permittivity of free space (a constant)
- A is the area of one of the plates
- d is the separation distance between the plates

Now, let's consider the initial condition where the plates are 1.0 mm apart.

Given:
- Potential difference (V₁) = 80 Volts
- Initial separation distance (d₁) = 1.0 mm = 0.001 m

From the formula above, we can write the equation as:

C₁ = ε₀ * (A / d₁)

Now, let's consider the new condition where the plates are pulled apart to a separation distance of 3.5 mm.

We need to find the potential difference (V₂) in this case.

We know that the charge (Q) on the capacitor plates remains the same since it cannot leak off. Therefore, we have:

Q = C₁ * V₁ = C₂ * V₂

Where:
- Q is the charge on the capacitor plates.
- C₁ is the initial capacitance.
- V₁ is the initial potential difference.
- C₂ is the final capacitance.
- V₂ is the final potential difference.

Since Q, C₁, and V₁ are known, we can rearrange the equation to solve for V₂:

V₂ = (C₁ * V₁) / C₂

Now, we need to find the final capacitance C₂ at a separation distance of 3.5 mm.

Given:
- Final separation distance (d₂) = 3.5 mm = 0.0035 m

Using the formula for capacitance, we can find C₂ as follows:

C₂ = ε₀ * (A / d₂)

Finally, we substitute the known values and calculate V₂:

V₂ = (C₁ * V₁) / C₂

By following this process, you should be able to find the potential difference between the plates when they are pulled to a new separation distance of 3.5 mm.