The circuit in the figure below consists of two identical, parallel metal plates connected to identical metal springs, a switch, and a 130-V battery. With the switch open, the plates are uncharged, are separated by a distance d = 7.65 mm, and have a capacitance C = 2.00 µF. When the switch is closed, the distance between the plates decreases by a factor of 0.500.
What is the question???
To answer this question, we need to calculate the new distance between the plates when the switch is closed.
Given:
- Original plate separation (d) = 7.65 mm
- Capacitance (C) = 2.00 µF
- Decrease factor (0.500)
We can use the formula for capacitance of parallel plates:
C = ε₀ * A / d
Where:
- C is the capacitance
- ε₀ is the permittivity of free space (a constant value)
- A is the area of the plates
- d is the distance between the plates
Since the area and permittivity remain constant, we can write the equation as:
C₁ / d₁ = C₂ / d₂
Where:
- C₁ is the initial capacitance (2.00 µF)
- d₁ is the initial plate separation (7.65 mm)
- C₂ is the final capacitance (2.00 µF)
- d₂ is the final plate separation (to be calculated)
Rearranging the equation, we can solve for d₂:
d₂ = (C₂ / C₁) * d₁
Substituting the given values:
d₂ = (2.00 µF / 2.00 µF) * 7.65 mm * 0.500
Simplifying the equation:
d₂ = 3.825 mm
Therefore, the new distance between the plates when the switch is closed is 3.825 mm.