PLEASE EXPLAIN
A charge moves a distance of 1.7 cm in the
direction of a uniform electric field having
a magnitude of 200 N/C. The electrical
potential energy of the charge decreases by
9.69095 × 10−19 J as it moves.
Find the magnitude of the charge on the
moving particle. (Hint: The electrical poten-
tial energy depends on the distance moved in
the direction of the field.)
Answer in units of C.
work done = force * distance = change in energy
Work = Q E d
9.69095*10^-19 = Q (200) (1.7*10^-2)
Q is positive because its potential energy decreases as it moves with the field (the field pushes it down the potential).
To find the magnitude of the charge on the moving particle, we can use the formula for electrical potential energy:
ΔPE = q * ΔV
where ΔPE is the change in electrical potential energy, q is the charge, and ΔV is the change in electric potential. In this case, the charge moves a distance of 1.7 cm (or 0.017 m) in the direction of a uniform electric field with a magnitude of 200 N/C.
Since the electrical potential energy decreases by 9.69095 × 10−19 J, we can substitute these values into the formula:
9.69095 × 10−19 J = q * ΔV
The electric potential difference, ΔV, can be calculated using the formula:
ΔV = Ed
where E is the electric field magnitude and d is the distance moved in the direction of the field. Substituting the given values:
ΔV = (200 N/C) * (0.017 m) = 3.4 V
Now we can solve for the charge, q, by rearranging the equation:
q = ΔPE / ΔV
q = (9.69095 × 10−19 J) / (3.4 V)
Calculating this expression:
q = 2.85556 × 10^-19 C
Therefore, the magnitude of the charge on the moving particle is 2.85556 × 10^-19 C.
To find the magnitude of the charge on the moving particle, we can use the relationship between electrical potential energy and the magnitude of the charge. The electrical potential energy (U) of a charged object in an electric field is given by the equation:
U = q * V
where q is the magnitude of the charge and V is the potential difference.
In this case, the electrical potential energy decreases by 9.69095 × 10−19 J as the charge moves a distance of 1.7 cm in the direction of the uniform electric field. Therefore, the change in electrical potential energy (∆U) is -9.69095 × 10−19 J.
The potential difference (∆V) is related to the magnitude of the electric field (E) and the distance moved (d) by the equation:
∆V = E * d
Plugging in the given values, we have:
∆V = (200 N/C) * (1.7 cm)
To find the magnitude of the charge (q), we can rearrange the first equation:
q = ∆U / ∆V
Plugging in the values, we have:
q = (-9.69095 × 10−19 J) / [(200 N/C) * (1.7 cm)]
To obtain the answer in units of Coulombs (C), we need to convert centimeters (cm) to meters (m) by dividing by 100:
q = (-9.69095 × 10−19 J) / [(200 N/C) * (1.7 cm / 100 cm/m)]
Finally, we can simplify the equation and calculate the magnitude of the charge:
q = (-9.69095 × 10−19 J) / [(200 N/C) * (0.017 m)]
q ≈ -8.1 × 10^-19 C
The magnitude of the charge on the moving particle is approximately 8.1 × 10^-19 C.