GSO002 Problem 12

1. Energy Conservation

When plate B is at position $d$ (plate separation), the capacitance is $C = \frac{\varepsilon S}{d}$ . Since the voltage is held at $V$ , the charge changes and the battery does work.

Comparing the state when released ( $d = d_0$ , velocity $= 0$ ) with the closest approach ( $d = d_{min}$ , velocity $= 0$ ):

Work by battery $=$ change in spring energy $+$ change in electrostatic energy:

(C_{min} - C_0)V^2 = \frac{1}{2} k (d_0 - d_{min})^2 + \frac{1}{2} (C_{min} - C_0)V^2

Simplifying:

\frac{1}{2} k (d_0 - d_{min})^2 = \frac{1}{2} (C_{min} - C_0)V^2

2. Deriving the Quadratic Equation for $d_{min}$

Substituting $C = \frac{\varepsilon S}{d}$ :

k (d_0 - d_{min})^2 = \varepsilon S V^2 \left( \frac{d_0 - d_{min}}{d_0 d_{min}} \right)

Since $d_{min} \neq d_0$ , divide both sides by $(d_0 - d_{min})$ :

k (d_0 - d_{min}) = \frac{\varepsilon S V^2}{d_0 d_{min}}

Rearranging into a quadratic in $d_{min}$ :

d_{min}^2 - d_0 d_{min} + \frac{\varepsilon S V^2}{k d_0} = 0

Using the quadratic formula, the two roots are:

d = \frac{d_0 \pm \sqrt{d_0^2 - 4\frac{\varepsilon S V^2}{k d_0}}}{2}

Since plate B starts at $d_0$ and moves toward plate A, $d_{min}$ is the first turning point — the larger root:

d_{min} = \frac{d_0 + \sqrt{d_0^2 - 4\frac{\varepsilon S V^2}{k d_0}}}{2}

3. Numerical Calculation

\varepsilon S V^2 = (8.5 \times 10^{-12}) \times (2.0) \times (4.0 \times 10^2)^2 = 2.72 \times 10^{-6} \text{ J m}

k d_0 = (1.0 \times 10^1) \times (1.7 \times 10^{-2}) = 0.17 \text{ N}

4\frac{\varepsilon S V^2}{k d_0} = 4 \times \frac{2.72 \times 10^{-6}}{0.17} = 64 \times 10^{-6} = 0.64 \times 10^{-4} \text{ m}^2

d_0^2 = (1.7 \times 10^{-2})^2 = 2.89 \times 10^{-4} \text{ m}^2

D = d_0^2 - 4\frac{\varepsilon S V^2}{k d_0} = 2.89 \times 10^{-4} - 0.64 \times 10^{-4} = 2.25 \times 10^{-4} \text{ m}^2

\sqrt{D} = 1.5 \times 10^{-2} \text{ m}

d_{min} = \frac{1.7 \times 10^{-2} + 1.5 \times 10^{-2}}{2} = \frac{3.2 \times 10^{-2}}{2} = 1.6 \times 10^{-2} \text{ m} = 16 \text{ mm}

Answer: 16

GSO002 Problem 12

Minimum Gap of a Movable Plate Against Non-Linear Electrostatic Force

Problem Statement

Constraints

Input Format

Solution

1. Energy Conservation

2. Deriving the Quadratic Equation for $d_{min}$

3. Numerical Calculation

Minimum Gap of a Movable Plate Against Non-Linear Electrostatic Force

Problem Statement

Constraints

Input Format

Solution

1. Energy Conservation

2. Deriving the Quadratic Equation for dmind_{min}dmin​

3. Numerical Calculation

2. Deriving the Quadratic Equation for $d_{min}$