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Answer:
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
Explanation:
For this exercise we must use conservation of energy
the electric potential energy is
U =
for the proton at x = -1 m
U₁ =
for the electron at x = 1 m
U₂ =
starting point.
Em₀ = K + U₁ + U₂
Em₀ =
final point
Em_f =
energy is conserved
Em₀ = Em_f
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})
\frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²( )
we substitute the values
½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ ) = 9 109 (1.6 10-19) ²(
)
2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( )
2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷
r₂² -1 = (4.443 10⁸)⁻¹
r2 =
r2 = 1 m
therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m
I attached the full question.
We know that for a parallel-plate capacitor the surface charge density is given by the following formula:
Where V is the voltage between the plates and d is separation.
Voltage is by definition:
Voltage is analog to the mechanical work done by the force.
Above formula is correct only If the field is constant, and we can assume that it is since no function has been given.
The charge density would then be:
Please note that elecric permittivity of air is very close to elecric permittivity of vacum, it is common to use them <span>interchangeably</span>.
We know that for a parallel-plate capacitor the surface charge density is given by the following formula:
Where V is the voltage between the plates and d is separation.
Voltage is by definition:
Voltage is analog to the mechanical work done by the force.
Above formula is correct only If the field is constant, and we can assume that it is since no function has been given.
The charge density would then be:
Please note that elecric permittivity of air is very close to elecric permittivity of vacum, it is common to use them <span>interchangeably</span>.