A moving particle encounters an external electric field that decreases its kinetic energy from 9970 eV to 6340 eV as the particle moves from position A to position B The electric potential at A is 55.0 V and that at B is 19.0 V Determine the charge of the particle Include the algebraic sign or with your answer

q = -1.61x10⁻¹⁷ C

Explanation:

The charge of the particle can be found using the definition of the work done by electric force:  

        (1)

Where:

q: is the charge

ΔV: is the difference in electric potential

The work is also equal to:

   (2)

Where:

and are the electric potential energy of the points A and B, respectively.

Now, by conservation of energy we have:

    (3)

Where:  

and are the kinetic energy of the points A and B, respectively.

Rearranging equation (3):  

Solving the above equation for q:

Therefore, the charge of the particle is -1.61x10⁻¹⁷ C.

I hope it helps you!

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explanation: