A moving particle encounters an external electric field that decreases its kinetic energy from 9160 eV to 7160 eV as the particle moves from position A to position B. The electric potential at A is -58.0 V, and that at B is +22.0 V. Determine the charge of the particle. Include the algebraic sign (+ or -) with your answer.
q*deltaV=deltaKE
take the change in KE (in ev), convert to Joules, then solve for q
deltaV = positive eighty v
deltaKE=final-initial (it will be negative, it lost KE), indicating it slowed, ie positive charge
To determine the charge of the particle, we can use the relationship between electric potential energy (U) and electric potential (V):
U = qV
where U is the change in electric potential energy, q is the charge of the particle, and V is the change in electric potential.
We are given the change in kinetic energy (ΔK) as the particle moves from position A to position B. The change in kinetic energy is related to the change in electric potential energy as:
ΔK = ΔU
Since the kinetic energy is decreasing, we have:
ΔK = (final kinetic energy) - (initial kinetic energy)
ΔK = 7160 eV - 9160 eV
ΔK = -2000 eV
The change in electric potential energy is equal to the negative of the change in kinetic energy:
ΔU = -ΔK
ΔU = 2000 eV
Now we can use the relationship between electric potential energy and electric potential to find the charge of the particle:
ΔU = qΔV
Substituting the values we know:
2000 eV = q (22.0 V - (-58.0 V))
2000 eV = q (80.0 V)
Simplifying this equation:
q = (2000 eV) / (80.0 V)
q = 25 eV/V
Therefore, the charge of the particle is 25 eV/V, and we include the algebraic sign (+ or -) based on the direction of the potential from A to B. Since the electric potential at B is positive (+22.0 V) and the electric potential at A is negative (-58.0 V), the potential is increasing from A to B, indicating that the particle is moving against the electric field. Hence, the charge of the particle is negative.
Therefore, the charge of the particle is -25 eV/V.