a cathode ray beam is moving from d negativ to d positive plate of a parallel plate consider charged to a potential V , with d as d distance btw d plates if d cathode ray particle has mass 'm' & charge q den d accn of d particle in d direction of motion is

ans) qV/(2md)

a cathode ray particle of mass m just remains in balanced state with in an electric field of intnsty E the charge of d prtcl is

ans]rt2[ mg/E]

To understand the behavior of a cathode ray particle moving between two charged plates, we need to analyze the forces acting on it and how these relate to its acceleration and balance in an electric field. Let's break this down step by step.

Acceleration of the Cathode Ray Particle

When a cathode ray particle, which has a mass \( m \) and charge \( q \), moves from the negative plate to the positive plate of a parallel plate capacitor, it experiences an electric field \( E \) created by the potential difference \( V \) across the plates. The distance between the plates is denoted as \( d \).

Electric Field and Force

The electric field \( E \) between two parallel plates can be expressed as:

• E = V/d

Here, \( V \) is the potential difference and \( d \) is the distance between the plates. The force \( F \) acting on the charged particle due to the electric field is given by:

• F = qE

Substituting the expression for \( E \), we get:

• F = q(V/d)

Finding Acceleration

According to Newton's second law, the acceleration \( a \) of the particle can be calculated using the formula:

• a = F/m

Substituting the expression for force, we have:

• a = (qV/d) / m

This simplifies to:

• a = qV/(md)

Thus, the acceleration of the cathode ray particle in the direction of motion is \( \frac{qV}{md} \).

Balanced State in an Electric Field

Now, let's consider the scenario where the cathode ray particle is in a balanced state within an electric field of intensity \( E \). In this case, the particle experiences a gravitational force \( F_g \) acting downwards, which can be expressed as:

• F_g = mg

For the particle to remain in a balanced state, the electric force acting upwards must equal the gravitational force acting downwards:

• qE = mg

Solving for Charge

We can substitute the expression for the electric field \( E \) into the equation:

• q(V/d) = mg

Rearranging this gives us:

• q = \frac{mgd}{V}

Thus, the charge of the particle in a balanced state within the electric field can be expressed as \( \frac{mgd}{V} \).

Summary of Key Points

In summary, we derived two important relationships for a cathode ray particle moving between charged plates:

• The acceleration of the particle is given by \( \frac{qV}{md} \).
• The charge of the particle in a balanced state within an electric field is \( \frac{mgd}{V} \).

These equations illustrate the interplay between electric forces and gravitational forces on charged particles, which is fundamental in understanding the behavior of cathode rays in electric fields.