ELECTROSTATICS BYU 42

Solution to Question 42

1. Physical Setup and Coordinates

The device acts as an electromechanical transducer. We define the following parameters:

S: Area of the metal plates (and rubber pad).
d: Original length of the rubber pad (uncompressed).
x: Displacement of the movable plate (compression).
k: Stiffness (spring constant) of the rubber pad.
ε₀: Permittivity of free space.
ε_r: Dielectric constant of the rubber.

When charged, the electrostatic attraction between the plates compresses the rubber pad until the spring force balances the electrostatic force.

Figure 1: Configuration of the device as a charge meter (top) and electrostatic voltmeter (bottom).

Part (a): Charge Meter Relation

The total charge $q$ to be measured is fed to the terminals. This charge distributes equally on the two plates of the device. Thus, the charge on each plate is $q/2$.

The electrostatic force $F_e$ between the plates (carrying charge $q/2$ each) is given by:

$$ F_e = \frac{(q/2)^2}{2 \varepsilon_0 S} = \frac{q^2/4}{2 \varepsilon_0 S} = \frac{q^2}{8 \varepsilon_0 S} $$

This electrostatic force is balanced by the restoring force of the rubber pad, $F_s = kx$.

$$ \frac{q^2}{8 \varepsilon_0 S} = kx $$

Solving for $q$:

$$ q^2 = 8 \varepsilon_0 S k x $$ $$ q = \sqrt{8 \varepsilon_0 S k x} $$

Part (b): Accuracy of Charge Meter

The accuracy depends on the rate of change of the reading ($x$) with respect to the quantity being measured ($q$), or conversely, the error propagation. From the handwritten solution, we analyze the derivative:

$$ q = \sqrt{8 \varepsilon_0 S k} \cdot x^{1/2} $$ $$ \frac{dq}{dx} \propto \frac{1}{\sqrt{x}} $$

For a fixed reading uncertainty $\Delta x$ (e.g., visual least count), the uncertainty in the measured charge $\Delta q$ is given by $\Delta q \approx \frac{dq}{dx} \Delta x$.

To minimize the relative error, we look for where $\frac{dq}{dx}$ is smaller. Since $\frac{dq}{dx}$ is inversely proportional to $\sqrt{x}$, the error decreases as $x$ increases. Thus, the device is more accurate when the displacement $x$ is larger.

Answer: In the higher portion of the range.

Part (c): Electrostatic Voltmeter Relation

When used as a voltmeter, a potential difference $V$ is applied across plates A and B. The device acts as a capacitor with capacitance $C$, where charges $+Q$ and $-Q$ are induced on the plates such that $Q = CV$.

The capacitance $C$ for the plate separation $(d-x)$ with dielectric $\varepsilon_r$ is:

$$ C = \frac{\varepsilon_0 \varepsilon_r S}{d – x} $$

The electrostatic attractive force is given by $F_e = \frac{Q^2}{2 \varepsilon_0 S}$. Substituting $Q = CV$:

$$ F_e = \frac{C^2 V^2}{2 \varepsilon_0 S} $$

Equating this to the spring force $kx$:

$$ \frac{C^2 V^2}{2 \varepsilon_0 S} = kx $$ $$ V^2 = \frac{2 \varepsilon_0 S k x}{C^2} $$ $$ V = \frac{1}{C} \sqrt{2 \varepsilon_0 S k x} $$

Substituting the expression for $C$:

$$ V = \left( \frac{d – x}{\varepsilon_0 \varepsilon_r S} \right) \sqrt{2 \varepsilon_0 S k x} $$ $$ V = \frac{d – x}{\varepsilon_r} \sqrt{\frac{2 \varepsilon_0 S k x}{(\varepsilon_0 S)^2}} $$ $$ V = \frac{d – x}{\varepsilon_r} \sqrt{\frac{2 k x}{\varepsilon_0 S}} $$

Part (d): Accuracy of Voltmeter

We analyze the sensitivity by checking how $V$ varies with $x$.

$$ V(x) \propto (d – x)\sqrt{x} = d\sqrt{x} – x^{3/2} $$

Differentiating with respect to $x$:

$$ \frac{dV}{dx} \propto \left( \frac{d}{2\sqrt{x}} – \frac{3}{2}\sqrt{x} \right) = \frac{d – 3x}{2\sqrt{x}} $$

At very low values of $x$, the term $\frac{1}{\sqrt{x}}$ makes the derivative $\frac{dV}{dx}$ very large. A large slope means that a small error in reading $x$ corresponds to a very large error in calculating $V$.

As $x$ increases (moving towards the higher portion of the range), the derivative $\frac{dV}{dx}$ decreases in magnitude, meaning the error $\Delta V$ for a fixed $\Delta x$ becomes smaller. Therefore, the accuracy improves as $x$ increases.

Answer: In the higher portion of the range.