Question

A uniform field of 2 kN//C is the x direction. A point charge = 3 mu C initially at rest at the origin is released. What is K.E. of this charge at x = 4m ? Also, calculate V (4m) - V (0).

Answer 1

Answer:

HELLO!

Explanation:

solution :- we know, Potential difference between two point a and b is given by

∆V = , Here E is electric field intensity.

Given, E = 2kN/C and we have to find potential from 0 to 4m

e.g., ∆V =

= -[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 .

Hence k.e is 24mj

Answer 2

Answer:

24mj is the correct answer

Explanation:

hope the answer will help you