Motion of a Charged Particle in a Uniform Electric Field: Comprehensive NEET Physics Notes

1. Motion of a Charged Particle in a Uniform Electric Field

When a charged particle enters a uniform electric field, it experiences a constant force that alters its motion. This force depends on the charge of the particle and the strength of the electric field, influencing its trajectory.

1.1 Fundamentals of Motion in a Uniform Electric Field

Force and Acceleration

A charged particle with charge $q$ and mass $m$ placed in a uniform electric field $\vec{E}$ experiences a force $\vec{F}$ given by: $\vec{F}=q\vec{E}$

According to Newton’s second law, the acceleration $\vec{a}$of the particle is: $\vec{a}=\frac{\vec{F}}{m}=\frac{q\vec{E}}{m}$ This shows that acceleration is directly proportional to the charge and the electric field strength but inversely proportional to the mass.

Direction of Motion

• Positive Charge: Accelerates in the direction of the electric field.
• Negative Charge: Accelerates opposite to the direction of the electric field.

Visual Aid:

Diagram: Include a labeled diagram showing a positively charged and a negatively charged particle experiencing forces in a uniform electric field, with the direction of the field indicated.

1.2 Trajectory in a Uniform Electric Field

The motion of a charged particle in a uniform electric field depends on its initial velocity and the direction of the electric field:

1. Parallel to the Electric Field:
2. • When a charged particle starts moving along the electric field lines, it experiences uniform acceleration.
   • Equations of motion apply: $v=u+at$ and $s=ut+\frac{1}{2}a{t}^2$, where $u$ is the initial velocity.
2. Perpendicular to the Electric Field:
3. • When the charged particle is moving perpendicular to the electric field, it follows a parabolic trajectory, similar to projectile motion.
   • The horizontal and vertical motions can be described as:
   • • Horizontal motion: Uniform velocity $v_x=v_0$ (constant)
     • Vertical motion: Accelerated motion under force $F=qE$, with acceleration $a_y=\frac{qE}{m}$.

Mnemonic:

"PPE: Positive Parallel, Negative Opposite" — Remember that Positive charges accelerate Parallel to the electric field, while Negative charges accelerate Opposite.

Real-life Application:

Television Tubes and Cathode Ray Oscilloscopes (CROs): Charged particles (electrons) are accelerated and deflected in a controlled manner using electric fields to create images on screens.

Quick Recap

• A charged particle experiences a force in a uniform electric field: $\vec{F}=q\vec{E}$.
• Positive charges accelerate in the direction of the electric field, while negative charges accelerate in the opposite direction.
• The motion can be linear (parallel to the field) or parabolic (perpendicular to the field).

Practice Questions

1. A proton enters a uniform electric field with an initial velocity of $5\times 10^5\text{m/s}$. Calculate the acceleration experienced by the proton if the electric field strength is $2\times 10^4\text{N/C}$.
2. An electron is placed in a uniform electric field of strength $1.5\times 10^3\text{N/C}$. Find the force acting on it.
3. A charged particle with charge $+2\mu C$ and mass $1\text{mg}$ is accelerated by a uniform electric field of $500\text{N/C}$. What is the acceleration of the particle?

Solutions

1. Acceleration of Proton: Given:
   $F=qE=1.6\times 10^{-19}\times 2\times 10^4=3.2\times 10^{-15}N$ $a=\frac{F}{m}=\frac{3.2\times 10^{-15}}{1.67\times 10^{-27}}\approx 1.92\times 10^{12}m/s^2$
2. • $q=1.6\times 10^{-19}C$
   • $E=2\times 10^{4}N/C$
   • $m_{proton}=1.67\times 10^{-27}kg$
2. Force on Electron: Given:
   $F=qE=-1.6\times 10^{-19}\times 1.5\times 10^3=-2.4\times 10^{-16}N$
3. • $q=-1.6\times 10^{-19}C$
   • $E=1.5\times 10^{3}N/C$
3. Acceleration of Charged Particle: Given:
   $F=qE=2\times 10^{-6}\times 500=1\times 10^{-3}N$ $a=\frac{F}{m}=\frac{1\times 1}{1\times 10^{-6}}$