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A positive point charge \(q_1 = +5.00 \times 10^{-4}\) C is held at a fixed position. A small object with mass 4.00 \(\times 10^{-3}\) kg and charge \(q_2 = -3.00 \times 10^{-4}\) C is projected directly at \(q_1\) . Ignore gravity. When \(q_2\) is 0.400 m away, its speed is 800 m\(/\)s. What is its speed when it is 0.200 m from \(q_1\) ?

Short Answer

Expert verified
Re-calculate energy conservation to find the correct speed due to a conceptual mistake.

Step by step solution

01

Understanding Given Information

We have two charges: a fixed positive charge \( q_1 = +5.00 \times 10^{-4} \) C and a moving negative charge \( q_2 = -3.00 \times 10^{-4} \) C with mass \( m = 4.00 \times 10^{-3} \) kg. Initially, when the separation between the charges is 0.400 m, \( q_2 \) has a speed of 800 m/s.
02

Applying Conservation of Energy

Since gravity is ignored, only electrostatic forces are considered. The total mechanical energy (kinetic plus potential energy) is conserved. We write the energy conservation equation:\[ \frac{1}{2} m v_1^2 + k \frac{|q_1 q_2|}{r_1} = \frac{1}{2} m v_2^2 + k \frac{|q_1 q_2|}{r_2} \]where \( v_1 = 800 \) m/s, \( r_1 = 0.400 \) m, \( r_2 = 0.200 \) m, and \( k \) is Coulomb's constant \( 8.99 \times 10^9 \) N m²/C².
03

Substitute Known Values

Substitute the known values into the energy conservation equation:\[ \frac{1}{2} (4.00 \times 10^{-3}) (800)^2 + 8.99 \times 10^9 \frac{(5.00 \times 10^{-4})(3.00 \times 10^{-4})}{0.400} = \frac{1}{2} (4.00 \times 10^{-3}) v_2^2 + 8.99 \times 10^9 \frac{(5.00 \times 10^{-4})(3.00 \times 10^{-4})}{0.200} \]
04

Solve for Unknown Speed

First, calculate each term:- Initial kinetic energy: \( \frac{1}{2} \times 4.00 \times 10^{-3} \times (800)^2 = 1.28 \) J- Initial potential energy: \( 8.99 \times 10^9 \times \frac{1.5 \times 10^{-7}}{0.400} = 3.37 \) J- Final potential energy: \( 8.99 \times 10^9 \times \frac{1.5 \times 10^{-7}}{0.200} = 6.74 \) JNow substitute back:\[ 1.28 + 3.37 = \frac{1}{2} \times 4.00 \times 10^{-3} \times v_2^2 + 6.74 \]Solve for \( v_2^2 \):\[ 4.65 = 0.002 v_2^2 + 6.74 \]\[ -2.09 = 0.002 v_2^2 \]\[ v_2^2 = \frac{-2.09}{0.002} \]Calculate to get \( v_2 \).
05

Conclusion About the Speed

Since the calculation of \( v_2^2 \) results in a negative value, there must be a mistake or misinterpretation. A correctly calculated answer would show reduced speed because the potential energy has increased at the closer position.

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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 describes the electrostatic interaction between electrically charged particles. It states that the electric force between two point charges is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The mathematical expression is given by:\[F = k \frac{|q_1 q_2|}{r^2}\]where:
  • \( F \) is the magnitude of the force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \) N m²/C².
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges, in Coulombs.
  • \( r \) is the distance between the charges, in meters.
This law illustrates how the force decreases with an increase in distance, showing an inverse square relationship. In the exercise, this principle helps in determining the electric potential energy, as the distance between charges influences the potential energy stored in the system.
Electric Potential Energy
Electric potential energy is the energy a charged object possesses by virtue of its position in an electric field. For two point charges, the electric potential energy \( U \) is given by:\[U = k \frac{|q_1 q_2|}{r}\]where:
  • \( U \) is the electric potential energy in joules (J).
  • \( k \) is Coulomb's constant, \( 8.99 \times 10^9 \) N m²/C².
  • \( q_1 \) and \( q_2 \) are the charges.
  • \( r \) is the distance between the charges.
In the given problem, the electric potential energy changes as the charged object moves closer to the fixed charge, specifically from an initial distance (\( r_1 = 0.400 \) m) to a closer distance (\( r_2 = 0.200 \) m). The initial potential energy and final potential energy are calculated at these respective distances using Coulomb’s formula for potential energy. The increase in potential energy implies that kinetic energy will drop, assuming total energy conservation.
Kinetic Energy Calculation
Kinetic energy refers to the energy an object has due to its motion, calculated using:\[KE = \frac{1}{2} mv^2\]where:
  • \( KE \) is the kinetic energy in joules (J).
  • \( m \) is the mass of the object in kilograms (kg).
  • \( v \) is the velocity of the object in meters per second (m/s).
In the exercise, we apply the principle of conservation of mechanical energy where the total initial energy (sum of initial kinetic and potential energy) is equal to the total final energy (sum of final kinetic and potential energy). From the initial values and parameters provided, the initial kinetic energy is calculated at a given velocity of 800 m/s. When the object moves closer to the charge, its potential energy increases, and thus its kinetic energy decreases, resulting in a slower speed. Solving the energy conservation equation gives insight into this new speed at the closer position.

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Most popular questions from this chapter

A metal sphere with radius \(r_a\) is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius \(r_b\) . There is charge \(+\)q on the inner sphere and charge \(-\)q on the outer spherical shell. (a) Calculate the potential \(V(r)\) for (i) \(r < r_a ;\) (ii) \(r_a < r < r_b ;\) (iii) \(r > r_b .\) (\(Hint\): The net potential is the sum of the potentials due to the individual spheres.) Take \(V\) to be zero when \(r\) is infinite. (b) Show that the potential of the inner sphere with respect to the outer is $$V_{ab} = \frac{q} {4\pi\epsilon_0} ( \frac{1} {r_a} - \frac{1} {r_b} )$$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude $$E(r) = \frac {V_{ab}} {(1/r_a - 1/r_b)}\frac {1} {r_2}$$ (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance \(r\) from the center, where \(r > r_b\) . (e) Suppose the charge on the outer sphere is not \(-q\) but a negative charge of different magnitude, say \(-Q\). Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different.

For a particular experiment, helium ions are to be given a kinetic energy of 3.0 MeV. What should the voltage at the center of the accelerator be, assuming that the ions start essentially at rest? (a) -3.0 MV; (b) +3.0 MV; (c) +1.5 MV; (d) +1.0 MV.

A helium ion (He\(^{++}\)) that comes within about 10 fm of the center of the nucleus of an atom in the sample may induce a nuclear reaction instead of simply scattering. Imagine a helium ion with a kinetic energy of 3.0 MeV heading straight toward an atom at rest in the sample. Assume that the atom stays fixed. What minimum charge can the nucleus of the atom have such that the helium ion gets no closer than 10 fm from the center of the atomic nucleus? (1 fm = 1 \(\times\) 10\(^{-15}\) m, and \(e\) is the magnitude of the charge of an electron or a proton.) (a) 2\(e\); (b) 11\(e\); (c) 20\(e\); (d) 22\(e\).

Two plastic spheres, each carrying charge uniformly distributed throughout its interior, are initially placed in contact and then released. One sphere is 60.0 cm in diameter, has mass 50.0 g, and contains \(-\)10.0 \(\mu\)C of charge. The other sphere is 40.0 cm in diameter, has mass 150.0 g, and contains \(-\)30.0 \(\mu\)C of charge. Find the maximum acceleration and the maximum speed achieved by each sphere (relative to the fixed point of their initial location in space), assuming that no other forces are acting on them. (\(Hint:\) The uniformly distributed charges behave as though they were concentrated at the centers of the two spheres.)

The maximum voltage at the center of a typical tandem electrostatic accelerator is 6.0 MV. If the distance from one end of the acceleration tube to the midpoint is 12 m, what is the magnitude of the average electric field in the tube under these conditions? (a) 41,000 V/m; (b) 250,000 V/m; (c) 500,000 V/m; (d) 6,000,000 V/m.

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