Question

A conducting sphere of radius $R = 1.0$ m is charged to a potential $V_i = 1000$ V. A thin metal disc of radius $r = 1.0$ cm mounted on an insulating handle is touched with the sphere making contact with one of its flat faces and then separated. After separation the disc is earthed and the process is repeated until the potential of the sphere becomes $V_f = 999$ V. Approximately how many times has this process been repeated?

• A. 10
• B. 20
• C. 30
• D. 40

Answer 1

Answer: (D)

40

Answer 2

Let $R$ be the radius of the sphere and $r$ be the radius of the disc. The capacitance of the conducting sphere is $C_s = 4 \pi \epsilon_0 R$.

Let $V_n$ be the potential of the sphere after the $n$-th repetition of the process. Initially, the potential of the sphere is $V_0 = V_i = 1000$ V.

When the disc is touched to the sphere, it samples the potential at the contact point. The charge transferred $\Delta Q$ is proportional to the potential $V$ and can be expressed as $\Delta Q = C'V$, where $C'$ is an effective capacitance.

The potential of the sphere after one contact is $V' = \frac{Q - \Delta Q}{C_s} = V - \frac{\Delta Q}{C_s} = V - \frac{C'V}{C_s} = V \left(1 - \frac{C'}{C_s}\right)$. Thus, each contact reduces the potential by a factor of $\left(1 - \frac{C'}{C_s}\right)$.

After $N$ repetitions, the potential is $V_f = V_i \left(1 - \frac{C'}{C_s}\right)^N$.

We have $V_i = 1000$ V and $V_f = 999$ V, so $\frac{999}{1000} = \left(1 - \frac{C'}{C_s}\right)^N$.

The key is to estimate $\frac{C'}{C_s}$. The charge density on the sphere is $\sigma = \frac{Q}{4 \pi R^2} = \frac{C_s V}{4 \pi R^2} = \frac{4 \pi \epsilon_0 R V}{4 \pi R^2} = \frac{\epsilon_0 V}{R}$.

The charge transferred to the disc might be proportional to its area multiplied by the charge density. So $\Delta Q \sim \sigma \times (\text{area of disc}) = \frac{\epsilon_0 V}{R} \times (\pi r^2)$. Thus, $C' V \sim \frac{\epsilon_0 V}{R} \pi r^2$, which gives $C' \sim \frac{\epsilon_0 \pi r^2}{R}$.

Then $\frac{C'}{C_s} \sim \frac{\epsilon_0 \pi r^2 / R}{4 \pi \epsilon_0 R} = \frac{r^2}{4 R^2}$.

Given $R = 1.0$ m and $r = 0.01$ m, we have $\frac{C'}{C_s} = \frac{(0.01)^2}{4 (1)^2} = \frac{0.0001}{4} = 0.000025$.

So, $\frac{999}{1000} = (1 - 0.000025)^N$, which means $0.999 = (0.999975)^N$.

Taking the natural logarithm of both sides, we have $\ln(0.999) = N \ln(0.999975)$. Thus, $N = \frac{\ln(0.999)}{\ln(0.999975)}$.

Using a calculator, $\ln(0.999) \approx -0.0010005$ and $\ln(0.999975) \approx -0.0000250003$.

Therefore, $N \approx \frac{-0.0010005}{-0.0000250003} \approx 40.0198 \approx 40$.

The number of repetitions is approximately 40.