Problem 16
Question
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 \(+3.00\) -n \(\mathrm{charge}\)?
Step-by-Step Solution
Verified Answer
The electron's speed is approximately \(2.42 \times 10^6 \text{ m/s}\) at 10 cm from the +3.00 nC charge.
1Step 1: Identify the Problem
We need to find the speed of an electron released midway between two point charges, before it reaches 10 cm away from the +3.00 nC charge.
2Step 2: Determine Initial and Final Positions
The midpoint is 25 cm from each charge. The final position of the electron is 10 cm from the +3.00 nC charge, which means it is 40 cm from the +2.00 nC charge.
3Step 3: Calculate the Electric Potential Energy Change
Use the formula for electric potential energy change: \[\Delta U = k \cdot \left(\frac{q_1 q_e}{r_{1i}} + \frac{q_2 q_e}{r_{2i}} - \frac{q_1 q_e}{r_{1f}} - \frac{q_2 q_e}{r_{2f}}\right)\] where: \(k = 8.99 \times 10^9 \ \text{N}\cdot\text{m}^2/\text{C}^2\),\(q_1 = +3.00 \text{ nC}\),\(q_2 = +2.00 \text{ nC}\),\(q_e = -1.60 \times 10^{-19} \text{ C} (electron's charge)\),\(r_{1i} = 0.25 \text{ m}\), \(r_{2i} = 0.25 \text{ m}\),\(r_{1f} = 0.10 \text{ m}\), \(r_{2f} = 0.40 \text{ m}\). Calculate the values and find \(\Delta U\).
4Step 4: Calculate the Kinetic Energy Change
Since the electron is released from rest, its initial kinetic energy is 0. The change in kinetic energy \(\Delta K\) will be equal to the negative change in potential energy, \(\Delta K = -\Delta U\).
5Step 5: Find the Final Speed
Using the equation for kinetic energy \(\Delta K = \frac{1}{2}mv^2\), solve for the speed \(v\) after substituting the change in kinetic energy from the potential energy change, and \(m = 9.11 \times 10^{-31} \text{ kg}\) (mass of an electron).
Key Concepts
Point ChargesKinetic EnergyElectron Motion
Point Charges
In the realm of physics, point charges are theoretical charges that are considered to be concentrated into a single point in space. Despite being idealizations, point charges help us simplify and solve complex electrostatic problems.
The exercise deals with two such point charges, one with a charge of +3.00 nC and the other with +2.00 nC. These point charges create an electric field that influences the motion of an electron placed between them. When an electron is at rest between point charges, its movement is dictated by the electric forces exerted by these charges.
The exercise deals with two such point charges, one with a charge of +3.00 nC and the other with +2.00 nC. These point charges create an electric field that influences the motion of an electron placed between them. When an electron is at rest between point charges, its movement is dictated by the electric forces exerted by these charges.
- Point charges exert a force on each other described by Coulomb's Law.
- The strength of this force depends on the magnitude of the charges and the distance between them.
- Coulomb's Law is expressed as:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where \( F \) is the force between the charges, \( k \) is Coulomb's constant, and \( r \) is the distance between the charges.
Kinetic Energy
Kinetic energy (KE) is the energy associated with the motion of an object. For an electron, which is a tiny particle, the kinetic energy tells us how fast it's moving. When we solve problems involving charged particles, understanding kinetic energy is crucial.
Initially, the electron in our exercise is at rest, which means its kinetic energy is zero. As the electron is released, it begins to move under the influence of the electric field created by the point charges. This electric field does work on the electron, converting electric potential energy into kinetic energy.
Initially, the electron in our exercise is at rest, which means its kinetic energy is zero. As the electron is released, it begins to move under the influence of the electric field created by the point charges. This electric field does work on the electron, converting electric potential energy into kinetic energy.
- The basic formula for kinetic energy is:
\[ KE = \frac{1}{2}mv^2 \]
where \( m \) is the mass of the electron and \( v \) is its velocity. - The change in kinetic energy, \( \Delta KE \), is significant as it helps us find the electron's speed at a given point.
- According to energy conservation, the change in kinetic energy is equal to the negative of the change in electric potential energy:
\[ \Delta KE = - \Delta U \]
Electron Motion
Electron motion in an electric field is a fascinating aspect of electromagnetism, describing how electrons travel among electric forces. In our context, an electron starts from rest midway between two point charges and is observed as it moves closer to one charge.
The motion of the electron is influenced by the electric field created by the point charges. The electron accelerates as it experiences a force due to the potential difference between the charges.
The motion of the electron is influenced by the electric field created by the point charges. The electron accelerates as it experiences a force due to the potential difference between the charges.
- The initial position is equidistant from the point charges, meaning the net force initially is zero, but as the electron starts to move, it gets influenced non-uniformly.
- The electron moves towards the +3.00 nC charge, losing electric potential energy while gaining kinetic energy.
- As the electron approaches the +3.00 nC charge to a position 10 cm from it, this change in position represents a change in its electric potential energy, which subsequently changes its speed.
Other exercises in this chapter
Problem 14
A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{\boldsymbol{E}}\) directed to the left. It is released from rest and mov
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A charge of 28.0 \(\mathrm{nC}\) is placed in a uniform electric field that is directed vertically upward and has a magnitude of \(4.00 \times 10^{4} \mathrm{V}
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Point charges \(q_{1}=+2.00 \mu \mathrm{C}\) and \(q_{2}=-2.00 \mu \mathrm{C}\) are placed at adjacent corners of a square for which the length of each side is
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Two charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charges have
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