A moving particle encounters an external electric field that decreases its kinetic energy from 9530 eV to 7800 eV as the particle moves from position A to position B. The electric potential at A is -69.0 V, and that at B is +39.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer.

ke change=9530-7800 figure that out in eV

charge*69=-kechange above.
divide both sides by 69, you get a charge in e.

multiply by e to get the charge in couloumbs.

To determine the charge of the particle, we can use the equation for the change in potential energy in an electric field:

ΔPE = qΔV

Where:
ΔPE = change in potential energy
q = charge of the particle
ΔV = change in electric potential

We are given:
ΔPE = 7800 eV - 9530 eV = -1730 eV (negative because the kinetic energy decreases)
ΔV = +39.0 V - (-69.0 V) = +108.0 V

Now, we need to convert the potential energy and electric potential to joules, as the SI unit for energy is the Joule (J), and 1 eV = 1.60 ✕ 10^(-19) J.

ΔPE = -1730 eV ✕ (1.60 ✕ 10^(-19) J/eV) = -2.768 ✕ 10^(-16) J
ΔV = +108.0 V

Substituting these values into the equation, we have:

-2.768 ✕ 10^(-16) J = q ✕ +108.0 V

Now, we can solve for q:

q = (-2.768 ✕ 10^(-16) J) / (108.0 V) = -2.565 ✕ 10^(-18) C

The charge of the particle is -2.565 ✕ 10^(-18) Coulombs.