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| | | ### Statement |
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| | − | $7.1.5.$ [Insert the problem statement] |
| | + | $7.1.5.$ A proton and an $\alpha$-particle, moving with the same velocity, enter a flat capacitor parallel to the plates. How many times greater will the deflection of the proton by the capacitor field from the rectilinear trajectory be than the deflection of the $\alpha$-particle? |
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| | | ### Solution |
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| | − | ### Statement |
| | + | We assume that the particles are only acted upon by the force from the electric field. It acts in the direction perpendicular to the plates. Since no forces act in the direction parallel to the plates, the momentum of the particles along this direction is conserved, and therefore, so is the velocity. |
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| | − | $7.1.4.$ A particle of mass $m$ with a charge $q > 0$ enters a flat capacitor, the plates of which are metal grids. The field strength in the capacitor is $E$, the distance between the grids is $d$. The initial velocity $v$ of the particle makes an angle $\alpha$ with the plane of the first grid. With what velocity and at what angle to the plane of the second grid will the particle fly out of the capacitor? |
| | + | We find the deflection in a uniform field from the kinematic formula: |
| | + | $$y = \frac{1}{2}at^2 \quad (1)$$ |
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| | − | ### Solution |
| | + | We find the acceleration from Newton's second law: |
| | + | $$a = \frac{F}{m} = \frac{qE}{m} \quad (2)$$ |
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| | − | The problem can be solved either using the kinematic approach or the energetic approach. The energetic method of solution is chosen here (it is shorter). |
| | + | If we denote $L$ as the length of the capacitor along the trajectory, and $v$ as the velocity along the capacitor plates, then the flight time is: |
| | + | $$t = \frac{L}{v} \quad (3)$$ |
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| | − | The work done by the electric field goes into changing the kinetic energy of the particle: |
| | − | $$\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = qEd$$ |
| | + | Substitute (2) and (3) into (1): |
| | + | $$y = \frac{1}{2} \cdot \frac{qE}{m} \cdot \left(\frac{L}{v}\right)^2 = \frac{qEL^2}{2mv^2}$$ |
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| | − | From here, we immediately find the magnitude of the velocity at which the particle exits the capacitor: |
| | − | $$u = v\sqrt{1 + \frac{2qEd}{mv^2}}$$ |
| | + | For a proton: $q = e$ and $m = m_p$. |
| | + | For an $\alpha$-particle (helium nucleus $\text{He}^{2+}$): $q = 2e$ and $m = 4m_p$. |
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| | − | Let's decompose the initial and final velocities of the particle along the axes: one axis is directed parallel to the plates (perpendicular to the field strength vector), the other axis is parallel to the field. |
| | − | Initial velocity components: $v_{\perp} = v \cos\alpha$ (this component does not change, since no forces act along this axis), $v_{\parallel} = v \sin\alpha$. |
| | + | Let's substitute these into the deflection formula: |
| | + | $$y_p = \frac{eEL^2}{2m_p v^2}$$ |
| | + | $$y_\alpha = \frac{2eEL^2}{2(4m_p)v^2} = \frac{eEL^2}{4m_p v^2}$$ |
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| | − | Final velocity components: $u_{\perp} = v_{\perp} = v \cos\alpha$, and we find the component $u_{\parallel}$ from the Pythagorean theorem: |
| | − | $$u_{\parallel} = \sqrt{u^2 - u_{\perp}^2} = \sqrt{\left(v^2 + \frac{2qEd}{m}\right) - v^2 \cos^2\alpha} = \sqrt{v^2 \sin^2\alpha + \frac{2qEd}{m}}$$ |
| | − | Or rewritten as: |
| | − | $$u_{\parallel} = v \sin\alpha \sqrt{1 + \frac{2qEd}{mv^2 \sin^2\alpha}}$$ |
| | + | Now let's find the ratio: |
| | + | $$\frac{y_p}{y_\alpha} = \frac{eEL^2 / (2m_p v^2)}{eEL^2 / (4m_p v^2)} = \frac{4}{2} = 2$$ |
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| | − | The tangent of the exit angle of the particle from the capacitor ($\beta$) is the ratio of the parallel component to the perpendicular component relative to the plates: |
| | − | $$\tan\beta = \frac{u_{\parallel}}{u_{\perp}} = \frac{v \sin\alpha \sqrt{1 + \frac{2qEd}{mv^2 \sin^2\alpha}}}{v \cos\alpha} = \tan\alpha \sqrt{1 + \frac{2qEd}{mv^2 \sin^2\alpha}}$$ |
| | + | The deflection of the proton is $2$ times greater than that of the $\alpha$-particle. |
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| | | #### Answer |
| | − | $$u = v\sqrt{1 + \frac{2qEd}{mv^2}}$$ |
| | − | $$\tan\beta = \tan\alpha \sqrt{1 + \frac{2qEd}{mv^2 \sin^2\alpha}}$$ |
| | − | |
| | − | #### Answer |
| | − | |
| | − | [Insert a concise answer or boxed result] |
| | + | By a factor of $2$ ($2$ times). |
