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.

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