In a plane capacitor with a uniform transverse electric field, a particle of charge e receives a transverse impulse $\Delta p_\perp = eE_\text{field}\,\Delta t$ The flight time is $\Delta t \approx L / v $
where L is the length of the plates and v is the initial velocity. For small angles, the deflection is:
where p is the initial momentum and v is the velocity. The same geometry and field imply$\alpha\propto 1/(p v)$
Calculation of p v for a particle with kinetic energy E
In relativity, the total energy is $E_{\text{total}} = E + mc^2$ The momentum satisfies$ (pc)^2 = E_{\text{total}}^2 - (mc^2)^2$ The velocity is$ v/c = pc / E_{\text{total}}$
Thus, the product pv is expressed in terms of E and the mass m:
$p v = \frac{E(E + 2mc^2)}{E + mc^2}$
Comparison of the angles
Since$\alpha\propto 1/(p v)$ the ratio between the deflection angle of the electron and that of the proton is the inverse of the ratio of their p v products:
The electron deflects approximately 50% more than the proton because, at the same kinetic energy, its smaller mass makes the product p v smaller, thereby increasing the deflection.
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The electron deflects approximately 50% more than the proton because, at the same kinetic energy, its smaller mass makes the product p v smaller, thereby increasing the deflection
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### Statement
### Statement
$14.4.16.$[Insert the problem statement]
$14.4.16.$
Flying through an electrostatic capacitor, a proton with kinetic energy E = 106
eV is deflected by an angle ap= 0.1 rad. Estimate the angle at which an
electron with the same kinetic energy will be deflected.
In a plane capacitor with a uniform transverse electric field, a particle of charge e receives a transverse impulse $\Delta p_\perp = eE_\text{field}\,\Delta t$ The flight time is $\Delta t \approx L / v $
In a plane capacitor with a uniform transverse electric field, a particle of charge e receives a transverse impulse $\Delta p_\perp = eE_\text{field}\,\Delta t$ The flight time is $\Delta t \approx L / v $
where L is the length of the plates and v is the initial velocity. For small angles, the deflection is:
where L is the length of the plates and v is the initial velocity. For small angles, the deflection is:
where p is the initial momentum and v is the velocity. The same geometry and field imply$\alpha\propto 1/(p v)$
where p is the initial momentum and v is the velocity. The same geometry and field imply$\alpha\propto 1/(p v)$
Calculation of p v for a particle with kinetic energy E
Calculation of p v for a particle with kinetic energy E
In relativity, the total energy is $E_{\text{total}} = E + mc^2$ The momentum satisfies$ (pc)^2 = E_{\text{total}}^2 - (mc^2)^2$ The velocity is$ v/c = pc / E_{\text{total}}$
In relativity, the total energy is $E_{\text{total}} = E + mc^2$ The momentum satisfies$ (pc)^2 = E_{\text{total}}^2 - (mc^2)^2$ The velocity is$ v/c = pc / E_{\text{total}}$
Thus, the product pv is expressed in terms of E and the mass m:
Thus, the product pv is expressed in terms of E and the mass m:
$p v = \frac{E(E + 2mc^2)}{E + mc^2}$
$p v = \frac{E(E + 2mc^2)}{E + mc^2}$
Comparison of the angles
Comparison of the angles
Since$\alpha\propto 1/(p v)$ the ratio between the deflection angle of the electron and that of the proton is the inverse of the ratio of their p v products:
Since$\alpha\propto 1/(p v)$ the ratio between the deflection angle of the electron and that of the proton is the inverse of the ratio of their p v products:
The electron deflects approximately 50% more than the proton because, at the same kinetic energy, its smaller mass makes the product p v smaller, thereby increasing the deflection.
The electron deflects approximately 50% more than the proton because, at the same kinetic energy, its smaller mass makes the product p v smaller, thereby increasing the deflection
