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Chapter 17: Problem 2316

Particle \(\mathrm{A}\) and \(\mathrm{B}\) have electric charge \(+\mathrm{q}\) and \(+4 \mathrm{q} .\) Both have mass \(\mathrm{m}\). If both are allowed to fall under the same p.d., ratio of velocities \(\left(\mathrm{V}_{\mathrm{A}} / \mathrm{V}_{\mathrm{B}}\right)=\ldots \ldots \ldots \ldots \ldots\) (A) \(2: 1\) (B) \(1: 2\) (C) \(1: 4\) (D) \(4: 1\)

Short Answer

Expert verified
The ratio of velocities \(\frac{V_{A}}{V_{B}}\) for Particle A and Particle B is \(1:2\).

Step by step solution

01

Consider the work-energy principle

The work-energy principle states that the work done on an object is equal to its change in kinetic energy. The work done on a charged particle in an electric field is determined by the charge, the potential difference, and the potential energy of the particle. We can find the kinetic energy of each particle after they fall through the potential difference and use it to calculate their velocities.
02

Write the equations for work done and kinetic energy

For particle A (charge +q): Work done, \(W_{A} = qV\) Change in kinetic energy, \(\Delta K_{A} = \frac{1}{2}mV_{A}^2\) For particle B (charge +4q): Work done, \(W_{B} = 4qV\) Change in kinetic energy, \(\Delta K_{B} = \frac{1}{2}mV_{B}^2\) Since work done is equal to the change in kinetic energy for each particle: \(qV = \frac{1}{2}mV_{A}^2\) and \(4qV = \frac{1}{2}mV_{B}^2\)
03

Solve for velocities

Now we need to solve for \(V_{A}\) and \(V_{B}\): For particle A: \(qV = \frac{1}{2}mV_{A}^2\) \(V_{A}^2 = \frac{2qV}{m}\) \(V_{A} = \sqrt{\frac{2qV}{m}}\) For particle B: \(4qV = \frac{1}{2}mV_{B}^2\) \(V_{B}^2 = \frac{8qV}{m}\) \(V_{B} = \sqrt{\frac{8qV}{m}}\)
04

Find the ratio of velocities

Now that we have expressions for \(V_{A}\) and \(V_{B}\), we can find the ratio between them: \(\frac{V_{A}}{V_{B}}=\frac{\sqrt{\frac{2qV}{m}}}{\sqrt{\frac{8qV}{m}}}= \sqrt{\frac{2}{8}}=\sqrt{\frac{1}{4}}\) \(\frac{V_{A}}{V_{B}}= \frac{1}{2}\) So the final answer is \(\frac{V_{A}}{V_{B}} = 1:2\) which corresponds to Option (B).

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of particles, symbolized as "+q" or "-q". These charges determine how particles interact in electric fields, either attracting or repelling each other. In the problem, particle A has a charge of "+q," while particle B has a charge of "+4q". This means particle B carries four times the charge of particle A.
This difference in charge is crucial because it determines how each particle responds to an electric field. It's like having two magnets, but one is much stronger than the other.
  • Positive charges repel positive charges.
  • Negative charges repel other negative charges.
  • Opposite charges attract each other.
Understanding electric charge helps us predict how charged particles will behave when exposed to a potential difference.
Potential Difference
Potential difference, often referred to as voltage, is the energy difference between two points in an electric field. It is a measure of how much work needs to be done to move a charge between these points.
In the exercise, both particles fall through the same potential difference \(V\). This means the energy available to each particle to do work (move) is the same, although how much energy they actually gain depends on their charge.
  • A higher potential difference means more energy for moving charges.
  • Calculated as the product of charge and potential difference (e.g., \(qV\) for work done).
This energy is then converted into kinetic energy, influencing how fast each particle moves.
Kinetic Energy
Kinetic energy is the energy that a particle has due to its motion. When particles fall through a potential difference, the work done on them converts into kinetic energy. This change in kinetic energy determines their speed.
The equations \(W_A = qV = \frac{1}{2}mV_A^2\) (for particle A) and \(W_B = 4qV = \frac{1}{2}mV_B^2\) (for particle B) show how this works:
  • Kinetic energy increases as particles gain speed.
  • Dependent on both mass and velocity: \(\frac{1}{2}mv^2\).
By setting work done equal to the change in kinetic energy, we can find the velocities of particles as they move through an electric field.
Velocity Ratio
The velocity ratio is a comparison of speeds between two particles. Here, it's determined by how much kinetic energy each particle gains. Given that the kinetic energy for each particle is derived from their respective charge and potential difference, we calculate their velocities and thus their ratio.
For particle A, \(V_A = \sqrt{\frac{2qV}{m}}\). For particle B, \(V_B = \sqrt{\frac{8qV}{m}}\). The velocity ratio \(\frac{V_A}{V_B} = \sqrt{\frac{1}{4}} = \frac{1}{2}\) tells us that particle A's velocity is half that of particle B.
  • A larger charge means more energy and usually more speed.
  • Velocity ratio illustrates the efficiency of energy conversion.
Understanding this ratio helps us predict relative motion between charged particles.

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Most popular questions from this chapter

For wave concerned with proton, de-Broglie wavelength change by \(0.25 \%\). If its momentum changes by \(\mathrm{P}_{\mathrm{O}}\) initial momentum \(=\ldots \ldots \ldots\) (A) \(100 \mathrm{P}_{\mathrm{O}}\) (B) \(\left\\{\mathrm{P}_{\mathrm{O}} / 400\right\\}\) (C) \(401 \mathrm{P}_{\mathrm{O}}\) (D) \(\left\\{\mathrm{P}_{\mathrm{O}} / 100\right\\}\)

An electron moving with velocity \(0.6 \mathrm{c}\), then de-brogly wavelength associated with is \(\ldots \ldots \ldots\) (rest mars of electron, \(\mathrm{m}_{0}=9.1 \times 10^{-31}(\mathrm{k} / \mathrm{s})\) \(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\) (A) \(3.24 \times 10^{-12} \mathrm{~m}\) (B) \(32.4 \times 10^{-12} \mathrm{~m}\) (C) \(320 \times 10^{-12} \mathrm{~m}\) (D) \(3.29 \times 10^{-14} \mathrm{~m}\)

In photoelectric effect, work function of martial is \(3.5 \mathrm{eV}\). By applying \(-1.2 \mathrm{~V}\) potential, photo electric current becomes zero, so......... (A) energy of incident photon is \(4.7 \mathrm{eV}\). (B) energy of incident photon is \(2.3 \mathrm{eV}\) (C) If photon having higher frequency is used, photo electric current is produced. (D) When energy of photon is \(2.3 \mathrm{eV}\), photo electric current becomes maximum

Wavelength of light incident on a photo-sensitive surface is reduced form \(3500 \AA\) to \(290 \mathrm{~mm}\). The change in stopping potential is \(\ldots \ldots . .\left(\mathrm{h}=6.625 \times 10^{-24} \mathrm{~J} . \mathrm{s}\right)\) (A) \(42.73 \times 10^{-2} \mathrm{~V}\) (B) \(27.34 \times 10^{-2} \mathrm{~V}\) (C) \(73.42 \times 10^{-2} \mathrm{~V}\) (D) \(43.27 \times 10^{-2} \mathrm{~V}\)

If ratio of threshold frequencies of two metals is \(1: 3\), ratio of their work functions is \(\ldots \ldots\) (A) \(1: 3\) (B) \(3: 1\) (C) \(4: 16\) (D) \(16: 4\)
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