Question

A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. What is the kinetic energy of this charge at x = 4m?

Answer 1

Your question -------> A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. What is the kinetic energy of this charge at x = 4m?

solution :- we know, Potential difference between two point a and b is given by
∆V = $-\bold{\int\limits^b_a\,E.dl}$, Here E is electric field intensity.

Given, E = 2kN/C and we have to find potential from 0 to 4m
e.g., ∆V = $-\bold{\int\limits^4_0\,(2kN/C)dl}$
= -[2 × 4] = -8kV = - 8 × 10³V

Now, potential energy , ∆U = q∆V
Charge , q is given 3μC = 3 × 10⁻⁶C
so, ∆U = 3 × 10⁻⁶ × (-8 × 10³) = -24 × 10⁻³ J

We know, according to conservation of energy theorem,
∆U + ∆K.E = 0
∆K.E = -∆U = -(-24 × 10⁻³) = 24 × 10⁻³ J

Hence, kinetic energy = 24 mJ