Question

The figure shows an infinite plane having uniform charge density $\sigma$ and a small charged particle having charge $q$ and mass $m$ suspended by a light insulating thread. Find $\sigma$ if the charge is in equilibrium.

• A. $\frac{2\epsilon_0 mg}{q}$
• B. $\frac{\epsilon_0 mg}{2q}$
• C. $\frac{2q}{\epsilon_0 mg}$
• D. $\frac{2q\epsilon_0}{mg}$

Answer 1

Answer: (A)

$\frac{2\epsilon_0 mg}{q}$

Answer 2

For an infinite plane with charge density $\sigma$, the electric field is

$$E = \frac{\sigma}{2\epsilon_0}.$$

The charged particle experiences an electric force

$$F_e = qE = \frac{q\sigma}{2\epsilon_0}.$$

Since the thread makes a $45^\circ$ angle with the vertical, its tension $T$ has components:

• Vertical: $T\cos45 = mg$
• Horizontal: $T\sin45 = \frac{q\sigma}{2\epsilon_0}$

Dividing the horizontal equilibrium by the vertical equilibrium:

$$\frac{T\sin45}{T\cos45} = \frac{\frac{q\sigma}{2\epsilon_0}}{mg} \quad \Rightarrow \quad \tan45 = \frac{q\sigma}{2\epsilon_0 mg}.$$

Since $\tan45=1$, we have:

$$\frac{q\sigma}{2\epsilon_0 mg} = 1 \quad \Rightarrow \quad \sigma = \frac{2\epsilon_0mg}{q}.$$