Answer:
To find the magnitude of the charge on the moving particle, we can use the formula for electric potential energy:
\[ \Delta U = q \cdot \Delta V \]
Where:
- \( \Delta U \) is the change in electrical potential energy,
- \( q \) is the charge of the particle, and
- \( \Delta V \) is the change in electric potential.
Since the particle moves in the direction of the electric field, the change in electric potential is related to the work done by the field on the particle:
\[ \Delta V = -E \cdot \Delta d \]
Where:
- \( E \) is the magnitude of the electric field, and
- \( \Delta d \) is the distance moved in the direction of the field.
Substituting this into the first equation, we get:
\[ \Delta U = -q \cdot E \cdot \Delta d \]
Given that \( \Delta U = 56.95 \times 10^{-19} \, \text{J} \), \( E = 224 \, \text{N/C} \), and \( \Delta d = 2.7 \, \text{cm} = 0.027 \, \text{m} \), we can solve for \( q \):
\[ 56.95 \times 10^{-19} \, \text{J} = -q \cdot (224 \, \text{N/C}) \cdot (0.027 \, \text{m}) \]
\[ q = \frac{56.95 \times 10^{-19} \, \text{J}}{224 \, \text{N/C} \cdot 0.027 \, \text{m}} \]
\[ q \approx 9.916 \times 10^{-19} \, \text{C} \]
So, the magnitude of the charge on the moving particle is approximately \( 9.916 \times 10^{-19} \, \text{C} \).
Answer:
9.4×10⁻¹⁹ C
Explanation:
Work done on the particle equals the change in energy.
W = ΔE
Fd = ΔE
(224 N/C) q (0.027 m) = 56.95×10⁻¹⁹ J
q = 9.4×10⁻¹⁹ C
