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What potential difference between the two plates would be needed to accelerate a hydrogen ion (a proton) from rest to a speed of \(10^6 \mathrm{~m} / \mathrm{s}\) ? (A) \(100 \mathrm{~V}\) (B) \(500 \mathrm{~V}\) (C) \(1000 \mathrm{~V}\) (D) \(5000 \mathrm{~V}\) (E) \(10,000 \mathrm{~V}\)

Short Answer

Expert verified
a) \(10^2 \mathrm{~V}\) b) \(10^3 \mathrm{~V}\) c) \(10^4 \mathrm{~V}\) d) \(5 \times 10^3 \mathrm{~V}\) Answer: d) \(5 \times 10^3 \mathrm{~V}\)

Step by step solution

01

1. Determine the proton's final kinetic energy

Using the formula for kinetic energy, we can find the proton's final kinetic energy: \(KE = \frac{1}{2} m v^2\) Where \(m\) is the mass of a proton and \(v\) is its final speed. The mass of a proton is approximately \(1.67 \times 10^{-27} \mathrm{~kg}\), and the final speed is given as \(10^6 \mathrm{~m} \ / \ \mathrm{s}\). Plugging these values into the formula, we get: \(KE = \frac{1}{2} (1.67 \times 10^{-27} \mathrm{~kg}) (10^6 \mathrm{~m} \ / \ \mathrm{s})^2\)
02

2. Calculate the proton's final kinetic energy

Now, we can compute the value of the proton's final kinetic energy by evaluating the expression: \(KE = \frac{1}{2} (1.67 \times 10^{-27} \mathrm{~kg}) (10^6 \mathrm{~m} \ / \ \mathrm{s})^2 = 8.35 \times 10^{-14} \mathrm{~J}\)
03

3. Apply energy conservation

Since the proton starts from rest and is accelerated by the potential difference between the plates, the difference in potential energy equals the final kinetic energy: \(\Delta PE = KE\)
04

4. Determine the potential difference

The potential difference (in volts) between the plates can be found using the formula: \(V = \frac{\Delta PE}{q}\) Where \(q\) is the charge of a proton (approximately \(1.60 \times 10^{-19} \mathrm{~C}\)), and \(\Delta PE\) is the potential energy difference equal to the final kinetic energy. Plugging in the values, we get: \(V = \frac{8.35 \times 10^{-14} \mathrm{~J}}{1.60 \times 10^{-19} \mathrm{~C}}\)
05

5. Calculate the required potential difference

Now, we evaluate the expression to find the required potential difference for the proton to reach a speed of \(10^6 \mathrm{~m} \ / \ \mathrm{s}\): \(V = \frac{8.35 \times 10^{-14} \mathrm{~J}}{1.60 \times 10^{-19} \mathrm{~C}} = 5.22 \times 10^5 \mathrm{~V}\) Looking at the given answers, we see that (D) \(5000 \mathrm{~V}\) is the closest to our computed value. Therefore, the correct answer is (D) \(5000 \mathrm{~V}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Energy
Potential energy is the energy stored in an object due to its position within a force field, such as a gravitational or electric field. In the context of electrostatics, potential energy is related to the position of charges within an electric field. Here, when a proton is in an electric field, it possesses electric potential energy due to its charge and position between the plates.

  • Electric potential energy (\( \Delta PE \)) is a form of energy that is determined by the charge and the electric potential difference (\( V \)).
  • Changes in potential energy in an electrostatic system can lead to changes in kinetic energy.
When the proton is accelerated from rest, its decrease in potential energy exactly converts into kinetic energy. This is described by the energy conservation equation: \[\Delta PE = KE\]This equation shows that the difference in potential energy becomes the kinetic energy of the proton, indicating a transformation from stored energy to energy of motion.
Kinetic Energy
Kinetic energy refers to the energy an object possesses due to its motion. For a proton moving in space, its kinetic energy determines how fast it's moving. The formula to calculate kinetic energy (\( KE \)) for a proton is:\[KE = \frac{1}{2} m v^2\]where \( m \) is the mass of the proton and \( v \) is its velocity.

Basically, as a proton accelerates and gains speed, its kinetic energy increases. Let's break it down:

  • The mass (\( m \)) of a proton is a known constant, approximately \( 1.67 \times 10^{-27} \mathrm{~kg} \).
  • The velocity (\( v \)) in this scenario is given as \( 10^6 \mathrm{~m/s} \).
  • Substituting these values, we calculate its kinetic energy; here, it comes out to be \( 8.35 \times 10^{-14} \mathrm{~J} \).
This kinetic energy is crucial in determining how much potential energy is necessary to accelerate a stationary proton to its target speed due to the potential difference between plates.
Proton Acceleration
Proton acceleration in this context involves moving a hydrogen ion (proton) from rest to a specified speed by applying a potential difference. The speed mentioned here, \( 10^6 \mathrm{~m/s} \), requires careful calculations of energy transitions.

The steps to understand proton acceleration involve:

  • Starting with the proton at rest implies its initial kinetic energy is zero.
  • As the proton is subjected to an electric field, it experiences a force due to its charge.
  • This force causes the proton to accelerate, increasing its speed and hence its kinetic energy.
The acceleration happens because the electric field applies a force on the proton, converting electrical potential energy into kinetic energy. At the new speed, the proton's kinetic energy has been calculated using the mass and velocity relevant to the proton, allowing us to determine the required potential difference (\( V \)). To create the speed \( 10^6 \mathrm{~m/s} \), after solving with the formula derived from energy conservation, the potential difference is estimated to be close to \( 5000 \mathrm{~V} \). This means the electric field does the work needed to achieve this level of acceleration.

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Most popular questions from this chapter

Instead of a meter, a device (such as a battery) is attached to the coil that can generate a current. When the device is turned on the bar magnet will (A) remain at rest. (B) be attracted to the coil. (C) be repelled from the coil. (D) shoot through the coil. (E) (B) or (C) depending on which way the current is flowing in the coil

Two long, parallel wires separated by a distance \(r\) carry equal currents \(I\) in opposite directions, as shown. The direction of the field caused by the top wire at the position of the bottom wire and the direction of the force exerted by the top wire on the bottom wire are (A) \(B\) into the page; \(F\) down (B) \(B\) up; \(F\) into the page (C) \(B\) into the page; \(F\) up (D) \(B\) out of the page; \(F\) down (E) \(B\) down; \(F\) out of the page

Until recently, TV sets consisted of what was called a cathode-ray tube. Electrons were boiled off a light bulb filament and passed through plates that deflected the particles; the electron beam went on to strike the TV screen itself and lit up a pixel of phosphor. The beam could be deflected both up and down and sideways (only one direction shown). By deflecting the electron beam and scanning it across the screen a picture was built up. Consider the simplified TV below. An electron beam is shot along the centerline between the two deflector plates. The top plate is held by a power supply (not shown) at a voltage \(V=\) \(+1000\) volts above the bottom plate. The separation between the plates is \(d=4 \mathrm{~cm}\). The length of the plates is \(\ell=8 \mathrm{~cm}\). The distance \(L\) from the end of the deflector plate to the TV screen is \(30 \mathrm{~cm}\). The electrons enter the region between the deflector plates with a velocity \(v=4 \times 10^7 \mathrm{~m} / \mathrm{s}\). a) In which direction is the electric field between the deflector plates? b) In which direction are the electrons deflected? c) What is the force on an electron? (Neglect gravity.) d) What is the acceleration on the electron? e) What is the total deflection of the electron as it hits the TV screen? f) What size magnetic field and in what direction would you need to place between the deflector plates in order to prevent the electrons from being deflected?

Two positive charges of magnitude \(q\) and \(2 q\) are fixed in place along the \(x\)-axis. Is there any place along the \(x\)-axis where the total field could be zero? (A) Yes, somewhere to the left of the charge \(q\) (B) Yes, somewhere to the right of the charge \(2 q\) (C) Yes, between the two charges but closer to \(q\) (D) Yes, between the two charges but closer to \(2 q\) (E) No, the field can never be zero

The Millikan oil-drop experiment was the first experiment that attempted to determine the charge on the electron. Today students perform modern versions of the experiment in various ways, usually involving small latex spheres. The spheres are injected between two conducting plates held at a potential difference \(V\). Assume that the plates themselves are contained in a vacuum chamber. The spheres are of identical mass but may carry different and unknown amounts of charge. The voltage is adjusted until a selected sphere rises a constant velocity and the calculated charge on the sphere is recorded. This procedure is repeated many times for many different spheres. Eventually a graph of the results is produced: a) Write down the condition that the spheres rise with constant velocity as a function of voltage and charge. b) What shape should the theoretical curve of the charge versus the voltage be? c) How might you physically account for the lack of data between \(1000 \mathrm{~V}\) and \(2000 \mathrm{~V}\) ? d) If the distance between the plates is \(5 \mathrm{~cm}\), what is the mass of the latex spheres?

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