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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 vf = (2m) q V + vi² 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 Work Done W = Change in Kinetic Energy EK

But since W = q V and EK = 12mv²:

q V = 12 m Δv² – 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, Δv =vf  vi 

Where 

vf is the final velocity and vi is the initial velocity

Substituting for Δv in Equation 1, we get:

q V = 12m (vf  vi)²

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

2m q V = vf²  vi²

vf² = 2m q V + vi² 

vf = 2m q V + vi²

Now, let’s substitute the values of m, q, V, and vi from the question in the above equation:

vf = 220 × 4 × 1 + 22

vf =25 + 4 

vf = 225

vf = 4.4

vf = 2.09 m/s

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


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