Electrostatics: Real-World Scenarios (6)

Question

A laser printer uses a charged drum to attract toner powder. The drum holds a surface charge density $\sigma = 5 \times 10^{-6}$ C/m$^2$. A toner particle of mass $m = 2 \times 10^{-12}$ kg and charge $q = -8 \times 10^{-15}$ C is released near the drum surface. Find the electric field just above the surface and the initial acceleration of the toner particle. (Take $\epsilon_0 = 8.85 \times 10^{-12}$ F/m.)

Solution — Step by Step

Final answer: $E \approx 5.65 \times 10^{5}$ N/C, $a \approx 2.26 \times 10^{3}$ m/s$^2$ towards the drum.

Why This Works

The factor $\sigma/\epsilon_0$ versus $\sigma/(2\epsilon_0)$ trips up most students. A conductor concentrates all its charge on the surface; the field inside the conductor is zero, and only points outward. Gauss’s law on a pillbox gives $E = \sigma/\epsilon_0$ outside.

For a thin sheet of charge in free space (no conductor), the field is $\sigma/(2\epsilon_0)$ on each side. Don’t mix the two cases.

Alternative Method

Apply Gauss’s law directly. Take a Gaussian pillbox spanning the drum surface. Flux through the outer face $= E \cdot A$. Inside the conductor, $E = 0$, so flux through the inner face $= 0$. Total enclosed charge $= \sigma A$. Then $E A = \sigma A / \epsilon_0 \Rightarrow E = \sigma / \epsilon_0$.

Common Mistake

Forgetting that the toner charge sign matters for direction, but only the magnitude matters for acceleration. A negative toner attracted to a positive drum still accelerates at $|q|E/m$ — the negative sign just flips the direction of motion.