A parallel plate capacitor with plates of area 320 cm.^2 is charged to a potential difference V and is then disconnected from the voltage source. When the plates are moved 0.3 cm. farther apart, the voltage between the plates is increased by 152 V. What is the charge on the positive plate of the capacitor?

Having trouble setting up equations to solve for the charge

To solve this problem, we can use the formula for the capacitance of a parallel plate capacitor:

C = ε₀ * (A/d)

where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation distance between the plates.

In this case, we know the initial area of the plates (320 cm²) and the change in distance (0.3 cm). We also know that the voltage between the plates increased by 152 V.

Let's break down the problem step by step:

Step 1: Convert the area from cm² to m².
320 cm² * (1 m / 100 cm)² = 0.032 m²

Step 2: Convert the separation distance from cm to m.
0.3 cm * (1 m / 100 cm) = 0.003 m

Step 3: Calculate the initial capacitance (C₁) using the formula.
C₁ = ε₀ * (A/d)
We need the permittivity of free space, which is ε₀ = 8.85 × 10⁻¹² F/m.
Plugging in the values:
C₁ = (8.85 × 10⁻¹² F/m) * (0.032 m² / 0.003 m)
C₁ ≈ 9.6 × 10⁻¹² F

Step 4: Calculate the final potential difference (V₂) using the given information.
The voltage between the plates increased by 152 V, so the final potential difference is V + 152 V.

Step 5: Calculate the final capacitance (C₂) using the formula.
C₂ = ε₀ * (A / (d + Δd))
We already know the permittivity of free space (ε₀) and the area (A), and we know the initial separation distance (d) and the change in distance (Δd).
C₂ = (8.85 × 10⁻¹² F/m) * (0.032 m² / (0.003 m + 0.3 cm))
C₂ ≈ 9.1 × 10⁻¹² F

Step 6: Now that we have both initial and final capacitance values, we can use the fact that the charge remains constant to set up an equation:
Q = C₁ * V = C₂ * (V + 152 V)

Step 7: Solve for Q, the charge on the positive plate.
To solve the equation, we'll substitute the values we know:
9.6 × 10⁻¹² F * V = 9.1 × 10⁻¹² F * (V + 152 V)

Simplifying the equation will give us the value of Q, the charge on the positive plate.