A moving particle encounters an external electric field that decreases its kinetic energy from 9100 eV to 6630 eV as the particle moves from position A to position B. The electric potential at A is -59.5 V, and the electric potential at B is +26.1 V. Determine the charge of the particle. Include the algebraic sign (+ or −) with your answer.
ΔKE=-ΔPE= q•Δφ.
9100-6630=2470 eV=2470•1.6 •10^-19 =3.95 •10^-16 J.
|Δφ|=|φ1- φ2|=|-59.5-26.1|=|-85.6|=85.6 V
q=3.95 •10^-16/85.6 = + 4.62•10^-18 C
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To find the charge of the particle, we can use the formula for the change in electric potential energy:
ΔPE = qΔV
where ΔPE is the change in electric potential energy, q is the charge of the particle, and ΔV is the change in electric potential.
Given that the kinetic energy of the particle decreases from 9100 eV to 6630 eV, we can find the change in electric potential energy as:
ΔPE = 9100 eV - 6630 eV
= 2470 eV
The change in electric potential, ΔV, is given by:
ΔV = VB - VA
= 26.1 V - (-59.5 V)
= 85.6 V
Substituting these values into the formula for the change in electric potential energy:
2470 eV = q * 85.6 V
Now we can solve for q:
q = (2470 eV) / (85.6 V)
= 28.85 e
Therefore, the charge of the particle is +28.85 e, where e is the elementary charge.
To determine the charge of the particle, we can use the equation that relates the change in electric potential energy (∆PE) to the charge of the particle (q) and the change in electric potential (∆V):
∆PE = q * ∆V
In this case, the change in electric potential is given by the difference between the electric potentials at positions A and B:
∆V = V_B - V_A = (+ 26.1 V) - (- 59.5 V) = 26.1 V + 59.5 V = 85.6 V
The change in kinetic energy (∆KE) of the particle can be calculated by subtracting the final kinetic energy (6630 eV) from the initial kinetic energy (9100 eV):
∆KE = KE_f - KE_i = 6630 eV - 9100 eV = -2470 eV
Now, we can substitute the given values into the equation:
-2470 eV = q * 85.6 V
To get the charge in terms of elementary charge (e), we can use the conversion factor:
1 eV = 1.6 x 10^-19 J
Therefore, we can rewrite the equation as:
-2470 e * 1.6 x 10^-19 J/eV = q * 85.6 V
Simplifying the equation gives:
-3.952 x 10^-16 J = q * 85.6 V
To isolate q, we divide both sides of the equation by 85.6 V:
q = -3.952 x 10^-16 J / 85.6 V
Calculating the result gives:
q ≈ -4.615 x 10^-18 C
Therefore, the charge of the particle is approximately -4.615 x 10^-18 Coulombs. The negative sign indicates that the particle has a negative charge.