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Two stationary point charges \(+3.00 \mathrm{nC}\) and \(+2.00 \mathrm{nC}\) are separated by a distance of 50.0 \(\mathrm{cm}\) . An electron is released from rest at a point midway between the two charges and moves along the line connecting the two charges. What is the speed of the electron when it is 10.0 \(\mathrm{cm}\) from the \(+3.00-\mathrm{charge}\) ?

Short Answer

Expert verified
The electron's speed is approximately 5.93 × 10^5 m/s.

Step by step solution

01

Identify Known Values and Constants

We have two charges, \(+3.00\,\mathrm{nC}\) and \(+2.00\,\mathrm{nC}\), separated by \(50.0\,\mathrm{cm}\). The electron is initially midway, and we want to find its speed \(v\) when it is \(10.0\,\mathrm{cm}\) away from the \(+3.00\,\mathrm{nC}\) charge. The charge of an electron, \(e\), is \(-1.60 \times 10^{-19}\,\mathrm{C}\), and its mass, \(m\), is \(9.11 \times 10^{-31}\,\mathrm{kg}\).
02

Calculate Initial Electric Potential Energy

Initially, the electron is at the midpoint. The electric potential \(V\) at the midpoint due to the charges is the sum of the potentials due to each charge:\[ V = k\left(\frac{q_1}{r_1} + \frac{q_2}{r_2}\right) \]where \(k = 8.99 \times 10^9\,\mathrm{N\,m^2\,/\,C^2}\), \(r_1 = r_2 = 0.25\,\mathrm{m}\), \(q_1 = 3.00 \times 10^{-9}\,\mathrm{C}\), and \(q_2 = 2.00 \times 10^{-9}\,\mathrm{C}\). This gives:\[ V = 8.99 \times 10^9 \left(\frac{3.00 \times 10^{-9}}{0.25} + \frac{2.00 \times 10^{-9}}{0.25}\right) \]
03

Compute Electric Potential Energy at Initial Point

The initial electric potential energy \(U_i\) is:\[ U_i = e \times V \]where \(e = 1.60 \times 10^{-19}\,\mathrm{C}\) (magnitude). Substitute the potential found in Step 2 to find \(U_i\).
04

Calculate Electric Potential at the New Position

Now, the electron is \(10.0\,\mathrm{cm}\) (or \(0.10\,\mathrm{m}\)) from the \(+3.00\,\mathrm{nC}\) charge, making it \(50.0 - 10 = 40.0\,\mathrm{cm}\) (or \(0.40\,\mathrm{m}\)) from the \(+2.00\,\mathrm{nC}\) charge. Calculate the potential \(V_f\) again:\[ V_f = k\left(\frac{3.00 \times 10^{-9}}{0.10} + \frac{2.00 \times 10^{-9}}{0.40}\right) \]
05

Compute Final Electric Potential Energy

The final electric potential energy \(U_f\) is:\[ U_f = e \times V_f \]Substitute the new potential from Step 4 here to find \(U_f\).
06

Apply Conservation of Energy

According to the conservation of energy:\[ U_i + K_i = U_f + K_f \] Initially, \(K_i = 0\), so:\[ U_i = U_f + \frac{1}{2}mv^2 \]Rearrange to solve for \(v\):\[ v = \sqrt{\frac{2(U_i - U_f)}{m}} \]
07

Calculate the Speed of the Electron

Using the values of \(U_i\) and \(U_f\) computed earlier, calculate the speed \(v\) of the electron. Substitute these values into the equation derived in Step 6 to obtain the final result for \(v\).

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle of electrostatics, describing the force between two point charges. It states that the electric force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:\[F = k\frac{|q_1q_2|}{r^2}\]where \(F\) is the force between the charges, \(k\) is Coulomb's constant \((8.99 \times 10^9 \, \text{N m}^2/\text{C}^2)\), \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the separation distance.
  • This law helps in calculating the effect of each charge on the electron in the problem.
  • It explains the interaction between the point charges and the moving electron.
This principle also allows us to determine the potential energy at any point, contributing to solving questions that involve multiple forces. In scenarios like ours, where an electron is between two charges, calculating the net force involves examining individual forces resulting from different charges. The calculations depend heavily on distance, making accurate values crucial.
Conservation of Energy
The Conservation of Energy principle states that the total energy in a closed system remains constant over time. It can neither be created nor destroyed but only transformed from one form to another. For our scenario, we focus on converting electric potential energy into kinetic energy.
  • The initial potential energy of the electron when it's midway is transformed into kinetic energy as the electron moves.
  • Using conservation equations, we compare initial and final total energies to solve for variables like velocity.
At the electron's initial position, all energy is in the form of electric potential energy \(U_i\). As the electron moves towards the charge, potential energy decreases while kinetic energy \(K\) increases. Thus, when applying conservation: \[U_i + K_i = U_f + K_f\]Since the electron starts from rest, \(K_i = 0\). Therefore, the equation simplifies to find the speed of the electron when considering changes in potential energy from its initial to final position.
Kinematics of Particles
Kinematics deals with the motion of objects without analyzing the forces causing the movement. In the context of charged particles like electrons, kinematics allows us to track their trajectory and speed based on energy transformations.
  • Once we have the energy changes, kinematics provides the mathematical framework to derive velocity.
  • By interpreting energy transformations, the speed at various points can be deduced.
For this problem, once the difference in potential energy is known using conservation laws, we employ kinematics to determine velocity. The kinematic equation derived specifically from the energy principle is:\[v = \sqrt{\frac{2(U_i - U_f)}{m}}\]This equation directly relates energy differences to motion parameters. Kinematics thus serves as a bridge from abstract energy concepts to practical calculations of motion characteristics like speed.

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Most popular questions from this chapter

A particle with charge \(+7.60 \mathrm{nC}\) is in a uniform electric field directed to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest, it moves to the right. After it has moved 8.00 \(\mathrm{cm}\) , the additional force has done \(6.50 \times 10^{-5} \mathrm{J}\) of work and the particle has \(4.35 \times 10^{-5} \mathrm{J}\) of kinetic energy. (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of the electric field?

Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides \(d\) . Two of the point charges are identical and have charge \(q\) . If zerv net work is required to place the three charges at the comers of the triangle, what must the value of the third charge be?

Two positive point charges, each of magnitude \(q,\) are fixed on the \(y\) -axis at the points \(y=+a\) and \(y=-a\) . Take the potential to be zero at an infinite distance from the charges. (a) Show the positions of the charges in a diagram. (b) What is the potenntial \(V_{0}\) at the origin? (c) Show that the potential at any point on the \(x\) -axis is $$ V=\frac{1}{4 \pi \epsilon_{0}} \frac{2 q}{\sqrt{a^{2}+x^{2}}} $$ (d) Graph the potential on the \(x\) -axis as a function of \(x\) over the range from \(x=-4 a\) to \(x=+4 a\) . (e) What is the potential when \(x \gg a ?\) Explain why this result is obtained.

Energy of the Nucleus. How much work is needed to assemble an atomic nucleus containing three protons (such as Be) if we model it as an equilateral triangle of side \(2.00 \times 10^{-13} \mathrm{m}\) with a proton at each vertex? Assume the protons started from very far away.

A point charge \(Q=+4.60 \mu \mathrm{C}\) is held fixed at the origin. A second point charge \(q=+1.20 \mu \mathrm{C}\) with mass of \(2.80 \times 10^{-4} \mathrm{kg}\) is placed on the \(x\) -axis, 0.250 \(\mathrm{m}\) from the origin. (a) What is the electric potential energy \(U\) of the pair of charges? (Take \(U\) to be zero when the charges have infinite scparation.) (b) The scond point charge is released from rest. What is its speed when its distance from the origin is (i) \(0.500 \mathrm{m} ;\) (ii) \(5.00 \mathrm{m} ;\) (iii) 50.0 \(\mathrm{m} ?\)

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