Ask Question
4 September, 06:53

A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particle moves from position A to position B. The electric potential at A is - 55.0 V, and the electric potential at B is + 27.0 V. Determine the charge of the particle. Include the algebraic sign ( + or - ) with your answer.

+5
Answers (1)
  1. 4 September, 07:18
    0
    Given Information:

    KEa = 9520 eV

    KEb = 7060 eV

    Electric potential = Va = - 55 V

    Electric potential = Vb = + 27 V

    Required Information:

    Charge of the particle = q = ?

    Answer:

    Charge of the particle = + 4.8x10⁻¹⁸ C

    Explanation:

    From the law of conservation of energy, we have

    ΔKE = - qΔV

    KEb - KEa = - q (Vb - Va)

    -q = KEb - KEa/Vb - Va

    -q = 7060 - 9520/27 - (-55)

    -q = 7060 - 9520/27 + 55

    -q = - 2460/82

    minus sign cancels out

    q = 2460/82

    Convert eV into Joules by multiplying it with 1.60x10⁻¹⁹

    q = 2460 (1.60x10⁻¹⁹) / 82

    q = + 4.8x10⁻¹⁸ C
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A moving particle encounters an external electric field that decreases its kinetic energy from 9520 eV to 7060 eV as the particle moves ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers