Solution: Iterative Discharge of a Sphere
Physics Principles
This problem involves charge sharing and surface charge density. A small metal disc touching a larger charged sphere acts like a “proof plane.” When it touches the sphere’s surface, it effectively becomes part of the sphere’s surface area. Upon removal, it carries away the charge that resided on the area it covered.
1. Analyzing the Charge Removal Process
Let the radius of the sphere be $R$ and the radius of the small disc be $r$. The sphere initially has a potential $V_i$ and charge $Q_i$.
The potential of a sphere is given by $V = \frac{kQ}{R}$. Therefore, charge is directly proportional to potential: $Q \propto V$.
When the thin disc (radius $r$) is placed flat on the sphere, it covers a patch of the sphere’s surface. Since the disc is metallic and thin, the charge resides on the outer surface. The area of the disc is $A_{disc} = \pi r^2$. The total surface area of the sphere is $A_{sphere} = 4\pi R^2$.
Assuming uniform charge distribution (constant surface charge density $\sigma$), the charge $\Delta q$ removed by the disc in one contact is proportional to the area it covers:
$$ \Delta q = \sigma \times (\text{Area of disc}) = \frac{Q_{current}}{4\pi R^2} \times (\pi r^2) $$ $$ \Delta q = Q_{current} \left( \frac{r^2}{4R^2} \right) $$2. Iterative Reduction of Charge
After one touch, the charge remaining on the sphere, $Q_1$, is:
$$ Q_1 = Q_i – \Delta q = Q_i – Q_i \left( \frac{r^2}{4R^2} \right) = Q_i \left( 1 – \frac{r^2}{4R^2} \right) $$After $n$ such contacts, the final charge $Q_f$ will be:
$$ Q_f = Q_i \left( 1 – \frac{r^2}{4R^2} \right)^n $$Since potential $V \propto Q$, we can write the same relation for potential:
$$ V_f = V_i \left( 1 – \frac{r^2}{4R^2} \right)^n $$3. Approximating for Small Change
We are given that the potential drop is very small ($1000\text{ V} \to 999\text{ V}$). This implies the term in the bracket is close to 1, or $n$ is not infinitely large, allowing us to use the binomial approximation $(1 – x)^n \approx 1 – nx$ for $x \ll 1$.
Let $x = \frac{r^2}{4R^2}$. Then:
$$ V_f \approx V_i (1 – nx) $$ $$ V_f = V_i – V_i n \left( \frac{r^2}{4R^2} \right) $$Rearranging to solve for $n$:
$$ V_i – V_f = V_i n \frac{r^2}{4R^2} $$ $$ n = \frac{V_i – V_f}{V_i} \cdot \frac{4R^2}{r^2} $$4. Calculation
Substitute the given values:
- $V_i = 1000\text{ V}$
- $V_f = 999\text{ V}$
- $R = 1.0\text{ m}$
- $r = 1.0\text{ cm} = 0.01\text{ m}$