An electron is released from rest at the negative plate of a parallel-plate capacitor.
If the distance across the plate is 2.0 mm and the potential difference across the plate is 6.0 V, with what velocity does the electron hit the positive plate? Given: – m
e
=9.1×10
−31
kg – e=1.6×10
−19
C
The Correct Answer and Explanation is :
To determine the velocity of an electron as it travels from the negative plate to the positive plate of a parallel-plate capacitor, we can use the concepts of electric potential energy and kinetic energy.
Given Data:
- Distance between plates, (d = 2.0 \, \text{mm} = 2.0 \times 10^{-3} \, \text{m})
- Potential difference, (V = 6.0 \, \text{V})
- Mass of the electron, (m_e = 9.1 \times 10^{-31} \, \text{kg})
- Charge of the electron, (e = 1.6 \times 10^{-19} \, \text{C})
Explanation:
- Electric Potential Energy: The work done on the electron as it moves through the potential difference can be calculated using the formula:
[
W = qV
]
where (q) is the charge of the electron and (V) is the potential difference. For an electron,
[
W = (-e)(V) = -1.6 \times 10^{-19} \times 6.0 = -9.6 \times 10^{-19} \, \text{J}
]
The negative sign indicates that the electron is moving from a region of lower potential to a higher potential. - Kinetic Energy: The work done on the electron is converted into kinetic energy (KE) when it reaches the positive plate:
[
KE = \frac{1}{2} mv^2
]
Since the electron starts from rest, all the work done on it becomes kinetic energy:
[
KE = -W = 9.6 \times 10^{-19} \, \text{J}
]
Setting the kinetic energy equal to the work done:
[
\frac{1}{2} mv^2 = 9.6 \times 10^{-19}
] - Solving for Velocity:
[
\frac{1}{2} (9.1 \times 10^{-31}) v^2 = 9.6 \times 10^{-19}
]
[
v^2 = \frac{2 \times 9.6 \times 10^{-19}}{9.1 \times 10^{-31}} \approx 2.11 \times 10^{12}
]
[
v \approx \sqrt{2.11 \times 10^{12}} \approx 1.45 \times 10^6 \, \text{m/s}
]
Conclusion:
The velocity of the electron as it hits the positive plate of the capacitor is approximately (1.45 \times 10^6 \, \text{m/s}). This result illustrates the significant effect of electric fields in accelerating charged particles, such as electrons, over relatively short distances. The potential difference provided a substantial energy source that transformed into kinetic energy, demonstrating fundamental principles of electrostatics and energy conservation in electric fields.