An electrometer consists of vertical metal bar at the top of which is attached at hinrod which gets deflected from the bar under the action of an electric charge(fig.). The reading are take non aquadrant graduated indegrees . The length of the rod is l and its massism. What will be the charge when the rod of such an electrometer is deflected through an angle . Make the following assumptions: (a) the charge on the electrometer is equally distributed between the bar & the rod (b) the charges are concentrated at point A on the rod & at point Bon the bar. fig is attached

To tackle this problem, we need to apply some fundamental principles of electrostatics and mechanics. We will analyze the forces acting on the rod of the electrometer when it is deflected at an angle \( a \). The goal is to derive an expression for the electric charge based on the given parameters and assumptions.

Understanding the Forces at Play

When the rod is deflected, it experiences a couple of forces: the gravitational force acting downward and the electrostatic force due to the electric charge acting horizontally. Let’s break down these forces for clarity.

• Gravitational Force (Fg): This force acts downward and can be calculated using the formula:

Fg = m * g

where \( m \) is the mass of the rod and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

• Electrostatic Force (Fe): The horizontal force due to the electric charge can be expressed using Coulomb's law, which states that the force between two point charges is given by:

Fe = k * (q1 * q2) / r^2

In our scenario, we can consider charges concentrated at point A on the rod and point B on the bar. The exact expression for the electrostatic force will depend on the charges and their distance apart.

Setting Up the Equilibrium Condition

At the equilibrium position when the rod is deflected at angle \( a \), the net torque about the pivot point (point A) must be zero. The torque due to the gravitational force and the electrostatic force must balance each other out.

• Torque due to Gravitational Force (τg): This is calculated as:

τg = Fg * (l/2) * sin(a)

Here, \( l/2 \) is the distance from point A to the center of mass of the rod, and \( sin(a) \) represents the vertical component of the force acting on the rod.

• Torque due to Electrostatic Force (τe): This can be represented as:

τe = Fe * (l) * sin(a)

where \( l \) is the full length of the rod.

Equating the Torques

Setting the two torques equal gives us:

Fg * (l/2) * sin(a) = Fe * l * sin(a)

We can simplify this equation by canceling \( sin(a) \) (assuming \( a \) is not zero) and rearranging:

Fg * (1/2) = Fe

(m * g) * (1/2) = k * (q1 * q2) / r^2

Calculating the Charge

Now, assuming the charges on the rod and bar are equal (let's denote them both as \( q \)), the equation simplifies to:

(m * g) / 2 = k * (q^2) / r^2

Rearranging for \( q \), we find:

q^2 = (m * g * r^2) / (2 * k)

q = sqrt((m * g * r^2) / (2 * k))

Final Thoughts

This equation tells us how to calculate the charge \( q \) on the electrometer based on its mass \( m \), the gravitational acceleration \( g \), the distance \( r \) between the charges, and the electrostatic constant \( k \). By plugging in the known values, you can determine the charge when the rod is deflected at angle \( a \).