Problem Analysis

We have a parallel plate capacitor where the upper plate is movable and attached to a spring with spring constant $k$. The lower plate is fixed.

• Initial separation (no voltage): $d$
• Area of plates: $A$
• Voltage applied: $V$

When voltage $V$ is applied, the plates attract. The upper plate moves down by a distance $x$. The new separation is $d – x$. We need to find the maximum voltage $V$ such that the plates do not touch (i.e., the system has a stable equilibrium position before collapsing).

Step-by-Step Derivation

1. Forces acting on the upper plate

Let the upper plate move down by $x$.

1. Spring Force ($F_s$): The spring is stretched by $x$. It pulls the plate up. $$ F_s = kx $$
2. Electrostatic Force ($F_e$): The capacitor plates attract each other. The capacitance at separation $(d-x)$ is $C = \frac{\varepsilon_0 A}{d-x}$. The energy stored is $U = \frac{1}{2} C V^2 = \frac{\varepsilon_0 A V^2}{2(d-x)}$. The magnitude of the force is $F_e = \left| \frac{dU}{dx} \right|$ or using the standard formula $F_e = \frac{1}{2} \varepsilon_0 A E^2$. $E = \frac{V}{d-x}$. $$ F_e = \frac{1}{2} \varepsilon_0 A \left(\frac{V}{d-x}\right)^2 = \frac{\varepsilon_0 A V^2}{2(d-x)^2} $$

2. Equilibrium Condition

At equilibrium, the upward spring force balances the downward electric force: $$ kx = \frac{\varepsilon_0 A V^2}{2(d-x)^2} $$ We can rearrange this to express $V^2$: $$ V^2 = \frac{2k}{\varepsilon_0 A} x (d-x)^2 $$

3. Condition for Maximum Voltage (Stability Limit)

We need to find the maximum $V$. This corresponds to maximizing the function $f(x) = x(d-x)^2$ with respect to $x$. Why? Because for a given $V$, if there are solutions for $x$, equilibrium exists. If $V$ is too high, the electric force (rising as $1/(d-x)^2$) will always exceed the spring force (rising linearly as $x$), causing collapse. The limit occurs when the $V(x)$ curve reaches its peak.

Let $y = x(d-x)^2$. Differentiate with respect to $x$: $$ \frac{dy}{dx} = (d-x)^2 + x \cdot 2(d-x)(-1) $$ $$ \frac{dy}{dx} = (d-x) [ (d-x) – 2x ] $$ $$ \frac{dy}{dx} = (d-x) (d – 3x) $$

For maximum $V$, set $\frac{dy}{dx} = 0$: The solutions are $x=d$ (plates touch, trivial min) and $x = d/3$. So, the critical displacement is $x = d/3$.

4. Calculate Maximum Voltage

Substitute $x = d/3$ back into the voltage equation: $$ V_{max}^2 = \frac{2k}{\varepsilon_0 A} \left(\frac{d}{3}\right) \left(d – \frac{d}{3}\right)^2 $$ $$ V_{max}^2 = \frac{2k}{\varepsilon_0 A} \left(\frac{d}{3}\right) \left(\frac{2d}{3}\right)^2 $$ $$ V_{max}^2 = \frac{2k}{\varepsilon_0 A} \cdot \frac{d}{3} \cdot \frac{4d^2}{9} $$ $$ V_{max}^2 = \frac{8 k d^3}{27 \varepsilon_0 A} $$

Taking the square root: $$ V_{max} = \sqrt{\frac{8 k d^3}{27 \varepsilon_0 A}} $$