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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

A large parallel plate capacitor has plate separation of \(1.00 \mathrm{cm}\) and plate area of \(314 \mathrm{cm}^{2} .\) The capacitor is connected across a voltage of \(20.0 \mathrm{V}\) and has air between the plates. How much work is done on the capacitor as the plate separation is increased to $2.00 \mathrm{cm} ?$
The electric potential at a distance of \(20.0 \mathrm{cm}\) from a point charge is \(+1.0 \mathrm{kV}\) (assuming \(V=0\) at infinity). (a) Is the point charge positive or negative? (b) At what distance is the potential $+2.0 \mathrm{kV} ?$
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.
Draw some electric field lines and a few equipotential surfaces outside a positively charged conducting cylinder. What shape are the equipotential surfaces?
A parallel plate capacitor is attached to a battery that supplies a constant voltage. While the battery is still attached, a dielectric of dielectric constant \(\kappa=3.0\) is inserted so that it just fits between the plates. What is the energy stored in the capacitor after the dielectric is inserted in terms of the energy \(U_{0}\) before the dielectric was inserted?
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