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Chapter 24: Problem 76

A proton traveling along the \(x\) -axis at a speed of \(1.0 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters the gap between the plates of a \(2.0-\mathrm{cm}-\) wide parallel plate capacitor. The surface charge distributions on the plates are given by \(\sigma=\pm 1.0 \cdot 10^{-6} \mathrm{C} / \mathrm{m}^{2}\). How far has the proton been deflected sideways \((\Delta y)\) when it reaches the far edge of the capacitor? Assume that the electric field is uniform inside the capacitor and zero outside.

Short Answer

Expert verified
Answer: The sideways deflection of the proton when it reaches the far edge of the capacitor is 4.32 x 10^-4 m.

Step by step solution

01

Calculate the electric field inside the capacitor

Using the relation between electric field (E) and surface charge density (σ), we can find the electric field inside the capacitor: \(E = \frac{\sigma}{\epsilon_0}\) where \(ε_0\) is the vacuum permittivity (\(8.85 \cdot 10^{-12} \mathrm{C}^2 / \mathrm{N}\mathrm{m}^2\)). \(\sigma = 10^{-6} \mathrm{C}/\mathrm{m}^2\) \(E = \frac{10^{-6} \mathrm{C}/\mathrm{m}^2}{8.85 \cdot 10^{-12} \mathrm{C}^2 / \mathrm{N}\mathrm{m}^2} = 1.13 \cdot 10^5 \mathrm{N}/\mathrm{C}\)
02

Determine the force acting on the proton

The force acting on the proton due to the electric field can be calculated using: \(F = qE\) where \(q\) is the charge of the proton (\(1.6 \cdot 10^{-19} \mathrm{C}\)). \(F = (1.6 \cdot 10^{-19} \mathrm{C})(1.13 \cdot 10^5 \mathrm{N}/\mathrm{C}) = 1.81 \cdot 10^{-14} \mathrm{N}\)
03

Calculate the acceleration of the proton in the y-direction

Using Newton's second law, we can find the acceleration of the proton in the y-direction: \(F = ma\) \(a = \frac{F}{m}\) where \(m\) is the mass of the proton (\(1.67 \cdot 10^{-27} \mathrm{kg}\)). \(a = \frac{1.81 \cdot 10^{-14} \mathrm{N}}{1.67 \cdot 10^{-27} \mathrm{kg}} = 1.08 \cdot 10^{13} \mathrm{m}/\mathrm{s}^2\)
04

Calculate the time taken for the proton to travel across the width of the capacitor

Using the initial speed \(v_x = 1.0 \cdot 10^6 \mathrm{m}/\mathrm{s}\) and the width of the capacitor \(d = 2.0 \mathrm{cm} = 0.02 \mathrm{m}\), we can find the time taken for the proton to travel across the width: \(t = \frac{d}{v_x} = \frac{0.02 \mathrm{m}}{1.0 \cdot 10^6 \mathrm{m}/\mathrm{s}} = 2.0 \cdot 10^{-8} \mathrm{s}\)
05

Calculate the sideways deflection (Δy) of the proton

Using the time \(t\) and the acceleration \(a\) found in the previous steps, we can calculate the sideways deflection (Δy) of the proton: Δy = \(\frac{1}{2}at^2\) Δy = \(\frac{1}{2}(1.08 \cdot 10^{13} \mathrm{m}/\mathrm{s}^2)(2.0 \cdot 10^{-8} \mathrm{s})^2 = 4.32 \cdot 10^{-4} \mathrm{m}\) Therefore, the sideways deflection of the proton when it reaches the far edge of the capacitor is \(4.32 \cdot 10^{-4} \mathrm{m}\).

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Most popular questions from this chapter

A spherical capacitor is made from two thin concentric conducting shells. The inner shell has radius \(r_{1}\), and the outer shell has radius \(r_{2}\). What is the fractional difference in the capacitances of this spherical capacitor and a parallel plate capacitor made from plates that have the same area as the inner sphere and the same separation \(d=r_{2}-r_{1}\) between plates?

A parallel plate capacitor with air in the gap between the plates is connected to a \(6.00-\mathrm{V}\) battery. After charging, the energy stored in the capacitor is \(72.0 \mathrm{~nJ}\). Without disconnecting the capacitor from the battery, a dielectric is inserted into the gap and an additional \(317 \mathrm{~nJ}\) of energy flows from the battery to the capacitor. a) What is the dielectric constant of the dielectric? b) If each of the plates has an area of \(50.0 \mathrm{~cm}^{2}\), what is the charge on the positive plate of the capacitor after the dielectric has been inserted? c) What is the magnitude of the electric field between the plates before the dielectric is inserted? d) What is the magnitude of the electric field between the plates after the dielectric is inserted?

A parallel plate capacitor consisting of a pair of rectangular plates, each measuring \(1.00 \mathrm{~cm}\) by \(10.0 \mathrm{~cm},\) with a separation between the plates of \(0.100 \mathrm{~mm},\) is charged by a power supply at a potential difference of \(1.00 \cdot 10^{3} \mathrm{~V}\). The power supply is then removed, and without being discharged, the capacitor is placed in a vertical position over a container holding de-ionized water, with the short sides of the plates in contact with the water, as shown in the figure. Using energy considerations, show that the water will rise between the plates. Neglecting other effects, determine the system of equations that can be used to calculate the height to which the water rises between the plates. You do not have to solve the system.

A capacitor consists of two parallel plates, but one of them can move relative to the other as shown in the figure. Air fills the space between the plates, and the capacitance is \(32.0 \mathrm{pF}\) when the separation between plates is \(d=0.500 \mathrm{~cm} .\) a) A battery with potential difference \(V=9.0 \mathrm{~V}\) is connected to the plates. What is the charge distribution, \(\sigma\), on the left plate? What are the capacitance, \(C^{\prime},\) and charge distribution, \(\sigma^{\prime},\) when \(d\) is changed to \(0.250 \mathrm{~cm} ?\) b) With \(d=0.500 \mathrm{~cm}\), the battery is disconnected from the plates. The plates are then moved so that \(d=0.250 \mathrm{~cm}\) What is the potential difference \(V^{\prime},\) between the plates?

Must a capacitor's plates be made of conducting material? What would happen if two insulating plates were used instead of conducting plates?
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