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## A charged particle (q = 8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is released from

Question

A charged particle (q = 8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is released from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference VB VA?

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Physics
6 months
2021-07-31T17:00:35+00:00
2021-07-31T17:00:35+00:00 1 Answers
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## Answers ( )

The potential difference will be

600 V.Explanation:Here the charged particle is said to exhibit a charge of

-8 mC.So when it moves from point A to point B, it gains a kinetic energy of4.8 J. Also, the charged particle is said to exhibit an electrostatic potential energy at points A and B.The particle will obey conservation of energy. So the sum of energy exhibited by the particle at point B will be equal to the sum of energy exhibited by the particle at point A.

At point A, the kinetic energy is zero and the electrostatic energy will be represented a

s U₁.Then, the total energy at point

A = U₁Similarly, at point B, the charged particle will exhibit kinetic energy as well as electrostatic potential

energy (U₂).Total energy exhibited by the charged particles at point B =

U₂ + K.E = U₂+4.8.Then, as per

law of conservation of energy:It is known that

electrostatic potential energyis theproduct of charge with the potential at that point.So, and

Then,

Then, V₂-V₁ will be 600 V.

Thus, the potential difference will be

600 V.