CBSEGrade 12PhysicsElectric Charges and Fields

Motion of Charged Particles in Electric Fields?

A uniform electric field of 2 × 10^4 N/C is directed along the positive x-axis. A particle with charge 6 × 10^-8 C is moving in this field with an initial velocity of 2 × 10^5 m/s. If the particle is released from rest at a distance of 1 m from the point where the electric field is measured, what will be its speed when it reaches the point of release?

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📌 CONCEPT: The motion of charged particles in an electric field is governed by the Lorentz force equation, which states that the force experienced by a charged particle is equal to the product of its charge, the electric field strength, and the direction of the field.

📐 RULE / FORMULA: According to the Lorentz force equation, the force (F) on a charged particle is given by F = qE, where q is the charge and E is the electric field strength.

💡 WORKED EXAMPLE: A charged particle with q = 6 × 10^-8 C is moving in a uniform electric field of strength E = 2 × 10^4 N/C. The particle is initially at rest at a distance of 1 m from the point where the electric field is measured. The force on the particle is F = qE = (6 × 10^-8 C) × (2 × 10^4 N/C) = 0.012 N. Since the force is constant and the particle starts from rest, its acceleration is a = F/m. Using the equation v^2 = u^2 + 2as, we can calculate the final velocity of the particle when it reaches the point of release.

⚠️ COMMON MISTAKE: Students often forget to consider the direction of the electric field and the charge of the particle, which can lead to incorrect calculation of the force and the final velocity.

16 Oct 26