Question

A pith ball of mass $m$ grams and charge $Q$ is suspended using a massless silk thread near a large charged conducting metal sheet of area $A$ and surface charge density $\sigma$. If the silk thread makes an angle $\theta$ with the metal sheet, then $\tan\theta$ is equal to:

• A. $\frac{Q\sigma}{2\varepsilon_0 mg}$
• B. $\frac{Q\sigma}{\varepsilon_0 mg}$
• C. $\frac{Q\sigma}{2A\varepsilon_0 mg}$
• D. $\frac{\varepsilon_0 mg}{Q\sigma}$

Hint:

Be careful not to confuse a charged conducting sheet ($E = \frac{\sigma}{\varepsilon_0}$) with a thin non-conducting infinite sheet ($E = \frac{\sigma}{2\varepsilon_0}$). The problem specifies a conducting metal plate, which establishes twice the field strength!

Answer 1

The Correct Option is B

Understanding the Concept: The uniform electric field ($E$) close to the surface of a charged conducting plate with local surface charge density $\sigma$ is given by Gauss's Law as: \[ E = \frac{\sigma}{\varepsilon_0} \] An object with charge $Q$ inside this field experiences a horizontal electrostatic force $F_e = QE$. For a suspended mass in static equilibrium, this force balances against the tension components of the supporting string.
Step 1: Analyze the forces acting on the ball.
Let the string make an angle $\theta$ with the vertical line parallel to the sheet face. The three acting forces are:
Downward gravity force $= mg$
Horizontal electrostatic repulsion force $= QE = \frac{Q\sigma}{\varepsilon_0}$
Tension force along the string $= T$
Resolving components in equilibrium: \[ T \cos\theta = mg \quad \cdots (1) \] \[ T \sin\theta = F_e = \frac{Q\sigma}{\varepsilon_0} \quad \cdots (2) \]
Step 2: Divide the component equations to solve for $\tan\theta$.
Dividing equation (2) by equation (1): \[ \frac{T \sin\theta}{T \cos\theta} = \frac{\left(\frac{Q\sigma}{\varepsilon_0}\right)}{mg} \implies \tan\theta = \frac{Q\sigma}{\varepsilon_0 mg} \]