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Chapter 17: Problem 63

A tiny hole is made in the center of the negatively and positively charged plates of a capacitor, allowing a beam of electrons to pass through and emerge from the far side. If \(40.0 \mathrm{V}\) are applied across the capacitor plates and the electrons enter through the hole in the negatively charged plate with a speed of \(2.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) what is the speed of the electrons as they emerge from the hole in the positive plate?

Short Answer

Expert verified
Answer: To find the final speed of the electron, follow these steps: 1. Calculate the work done by the electric field: Work = q * V = \((-1.6 \times 10^{-19} C) * (40.0 V)\) 2. Calculate the change in kinetic energy: ΔKE = Work 3. Find the initial kinetic energy: Initial KE = \(\frac{1}{2} * (9.11 \times 10^{-31} kg) * (2.5 \times 10^6 m/s)^2\) 4. Find the final kinetic energy: Final KE = Initial KE + ΔKE 5. Calculate the final speed of the electron: \(v_f = \sqrt{\frac{2 * Final_KE}{(9.11 \times 10^{-31} kg)}}\) By following these steps, you will find the final speed of the electron as it emerges from the hole in the positive plate.

Step by step solution

01

Calculate the work done by the electric field

The work done by the electric field can be calculated by multiplying the charge of the electron with the potential difference across the plates: Work = q * V where q is the charge of an electron (\(q = -1.6 \times 10^{-19} C\) ) and V is the voltage (\(V = 40.0 V\)).
02

Calculate the change in the kinetic energy

The work done on an electron due to the electric field is equal to the change in its kinetic energy. As the electron's direction of motion remains unchanged, we can write: ΔKE = Work
03

Find the initial kinetic energy

The initial kinetic energy of an electron can be found using its initial speed: Initial KE = \(\frac{1}{2} * m_e * v_i^2\) where \(m_e = 9.11 \times 10^{-31} kg\) is the mass of an electron and \(v_i = 2.5 \times 10^6 m/s\) is its initial speed.
04

Find the final kinetic energy

Using the change in kinetic energy from Step 2, we can find the final kinetic energy of an electron: Final KE = Initial KE + ΔKE
05

Calculate the final speed of the electron

We can use the final kinetic energy to calculate the final speed of an electron using the following formula: Final KE = \(\frac{1}{2} * m_e * v_f^2\) Now, solving for \(v_f\), we get: \(v_f = \sqrt{\frac{2 * Final_KE}{m_e}}\) By following these steps, we will find the final speed of the electrons as they emerge from the hole in the positive plate.

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Most popular questions from this chapter

Capacitors are used in many applications where you need to supply a short burst of energy. A \(100.0-\mu \mathrm{F}\) capacitor in an electronic flash lamp supplies an average power of \(10.0 \mathrm{kW}\) to the lamp for $2.0 \mathrm{ms}$. (a) To what potential difference must the capacitor initially be charged? (b) What is its initial charge?
A spherical conductor with a radius of \(75.0 \mathrm{cm}\) has an electric field of magnitude \(8.40 \times 10^{5} \mathrm{V} / \mathrm{m}\) just outside its surface. What is the electric potential just outside the surface, assuming the potential is zero far away from the conductor?
A positively charged oil drop is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced $16 \mathrm{cm}$ apart. If the mass of the drop is \(1.0 \times 10^{-13} \mathrm{kg}\) and it remains stationary when the potential difference between the plates is $9.76 \mathrm{kV},$ what is the magnitude of the charge on the drop? (Ignore the small buoyant force on the drop.)
A parallel plate capacitor is connected to a battery. The space between the plates is filled with air. The electric field strength between the plates is \(20.0 \mathrm{V} / \mathrm{m} .\) Then, with the battery still connected, a slab of dielectric \((\kappa=4.0)\) is inserted between the plates. The thickness of the dielectric is half the distance between the plates. Find the electric field inside the dielectric.
A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\) The parallel plates are separated by \(0.40 \mathrm{mm}\) of bakelite. What is the capacitance of this capacitor?
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