1. Understanding the Physical Setup

An electron gun emits electrons away from the probe. By conservation of charge, as negative electrons leave the probe, the probe itself accumulates a net positive charge. This positive charge creates a positive electric potential $V$ on the sphere.

As the potential $V$ increases, it becomes harder for subsequent electrons to escape because the positive sphere attracts the negative electrons back. The emission stops (or rather, electrons are pulled back) when the work done to escape the potential equals the initial kinetic energy of the electrons.

2. Analyzing the Steady State

Let’s determine if the sphere reaches its maximum potential within the given time $t = 1.0$ min.

The maximum potential $V_{max}$ is reached when the electric potential energy magnitude equals the kinetic energy $W$ of the electrons:

Given $W = 9.0 \times 10^4 \text{ eV}$, the stopping potential is $V_{max} = 9.0 \times 10^4 \text{ V}$.

Let’s check the time required to reach this potential. The capacitance of an isolated sphere is $C = 4\pi\epsilon_0 r$. The charge required is $Q = C V_{max}$.

$$Q = \frac{r V_{max}}{k} = \frac{1.0 \times 9.0 \times 10^4}{9 \times 10^9} = 10^{-5} \text{ C}$$

The current is $I = 1.0 \text{ mA} = 10^{-3} \text{ C/s}$. The time to charge is:

$$t_{charge} = \frac{Q}{I} = \frac{10^{-5}}{10^{-3}} = 0.01 \text{ s}$$

Since the elapsed time ($1.0 \text{ min} = 60 \text{ s}$) is much larger than $0.01 \text{ s}$, the sphere has long reached its steady-state maximum potential $V_{max}$.

3. Calculating the Electric Field

The electric field strength $E$ at the surface of a charged sphere is related to its potential $V$ by the relationship:

Substituting the limiting potential $V_{max} = \frac{W}{|e|}$:

Plugging in the numerical values:

• $W/|e| = 9.0 \times 10^4 \text{ V}$ (Since energy in eV corresponds numerically to volts)
• $r = 1.0 \text{ m}$

$$E = \frac{9.0 \times 10^4 \text{ V}}{1.0 \text{ m}} = 9.0 \times 10^4 \text{ V/m}$$