How Do You Find the Final Velocity of a Charged Particle Moving with an Initial Velocity of 2 Meters per Second and Accelerated Through a Potential Difference of 1V? The Particle Has a Mass of 20 kg and a Charge of 4C.

Answer – The final velocity of the charged particle can be found using the formula $v_{f} = \sqrt{(\frac{2}{m}) q V + v_{i} ²}$ as 2.09 m/s.

Explanation:

When a charged particle moves through an electric field, work is done on it, causing its velocity to increase and hence its kinetic energy to change.

This work done on the charged particle is equal to the change in the kinetic energy of the particle.

So $W o r k D o n e W = C h a n g e i n K i n e t i c E n e r g y ∆ E_{K}$

But since $W = q V a n d ∆ E_{K} = 12 m ∆ v ²$ :

$q V = \frac{1}{2} m Δ v ²$ – Equation 1

Where

q is the charge of the particle

V is the potential difference it is accelerated through

m is the mass of the charged particle

Δv is the change in its velocity

Further, $Δ v = v_{f} - v_{i}$

Where

$v_{f}$ is the final velocity and $v_{i}$ is the initial velocity

Substituting for Δv in Equation 1, we get:

$q V = \frac{1}{2} m (v_{f} - v_{i}) ²$

This expression can be further simplified to get the formula for the final velocity of the particle:

$\frac{2}{m} q V = v_{f} ² - v_{i} ²$

$v_{f} ² = \frac{2}{m} q V + v_{i} ²$

$v_{f} = \sqrt{\frac{2}{m} q V + v_{i} ²}$

Now, let’s substitute the values of m, q, V, and $v_{i}$ from the question in the above equation:

$v_{f} = \sqrt{\frac{2}{20} \times 4 \times 1 + 2^{2}}$

$v_{f} = \sqrt{\frac{2}{5} + 4}$

$v_{f} = \sqrt{\frac{22}{5}}$

$v_{f} = \sqrt{4.4}$

$v_{f} = 2.09 m / s$

So the final velocity of the charged particle is found to be 2.09 m/s.

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