A moving particle encounters an external electric field that decreases its kinetic energy from 10360 eV to 7680 eV as the particle moves from position A to position B. The electric potential at A is -52.5 V, and the electric potential at B is +29.7 V. Determine the charge of the particle. Include the algebraic sign (+ or −) with your answer.
To determine the charge of the particle, we need to use the relationship between the change in electric potential energy (ΔPE) and the charge (q) of the particle.
ΔPE = q * ΔV
Where:
ΔPE = change in electric potential energy
q = charge of the particle
ΔV = change in electric potential
First, let's calculate the change in electric potential (ΔV) between positions A and B:
ΔV = Vb - Va
ΔV = (+29.7 V) - (-52.5 V)
ΔV = +82.2 V
Next, we substitute the given values into the equation for the change in electric potential energy (ΔPE):
ΔPE = q * ΔV
(7680 eV - 10360 eV) = q * (+82.2 V)
We need to convert the electric potential energy from electron-volt (eV) to joules (J), since the units must be consistent. 1 eV is equivalent to 1.6 x 10^-19 J.
(7680 eV - 10360 eV) * (1.6 x 10^-19 J/eV) = q * (+82.2 V)
Now, we solve for q:
-26880 * (1.6 x 10^-19 J/eV) = q * 82.2 V
Finally, we can calculate the charge of the particle:
q = (-26880 * 1.6 x 10^-19 J/eV) / 82.2 V
By performing the calculation, we get:
q ≈ -5.24 x 10^-19 C
Therefore, the charge of the particle is approximately -5.24 x 10^-19 C (negative sign indicates a negative charge).