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Ch 23: Electric Potential

Chapter 23, Problem 23

(a) How much work would it take to push two protons very slowly from a separation of 2.00x10^-10 m (a typical atomic distance) to 3.00x10^-15 m (a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?

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Hey, everyone today, we're dealing with the problem regarding work, electric potential energy and kinetic energy. So we're being told that in an attempt to merge two hydrogen ions into one single atom, the separation of the ions has changed slowly from 1.1 times 10 to the negative m, 22 times 10 to the negative 15 m. With this, we're being asked to do two things. The task in this problem is to determine a the work required in changing the electrical potential energy such that it causes the separation between the islands and also being asked to determine the speed at which the islands are moving after they've reached the initial separation, The initial separation of 1.1 times 10 to the negative 11th m. So we should understand that ions gain speed after release since the potential energy converts to kinetic energy and that will be utilized in the second part of the problem. But first, let's look at part a part. Hey, in part a right. So we need to find the work being done. Now what is work work is defined as the change in potential energy which in this case is electric potential energy. Now, potential energy is given by K yeah, constant multiplied by the two charges involved divided by R the distance between the charges. Now the charge of a hydrogen ion do like this queue of a hydrogen ion is equal to 1.6 times to the negative 19th columns. And we'll need that in a second. But why this is important is because in our specific example, this means it's the final potential energy minus D in this potential energy positions uh initial and final, right? And let me write that a little nicer, we can say that this is the initial position, initial and this is our final position. That's final and initial. But the only charges here are two hydrogen ions, which means Q one, Which means Q one, Q one, Excuse me is equal to Q two, which means they're both equal to E positive, which means we can rewrite this again as K E squared because we have Q one, Q two. And since both are here, that's E times eerie E squared over our final minus K E squared over our initial, we can factor out the E or the K ee rather K squared. And we get K squared into one over the final distance minus one over the initial distance. Since we're given those values, we can say this is our two or sorry, our initial and our final, we can just substitute all of this in and we get nine times 10 to the ninth Newton meter squared over Coolum square that is cool is constant K multiplied by 1.6 times 10 to the native 19th columns multiplied by one over one, excuse me, two times two times 10 to the negative 15th m minus one over 1.1 .1 times 10 to the negative 11th m which will give us an answer of, of 1.15 times 10 to the negative 13th Jules. So that's part A. Now let's look at part B oops. So part B is asking us to find the speed when the separation is at 1.1 times 10 to the negative 11 m. So they have the same masses, which means they'll have equal speeds, they'll have equal momentum. And this also implies that they will have equal kinetic energy. So the change let me write this out in red or no, we'll still stay with the change in potential energy here will therefore be equal to the sum of the kinetic energies after the hydrogen ions are released, right? Because we have to deal with inter conversion of or conservation of energy, right, which means that the kinetic energies that are resultant of the potential energy or that some of those kinetic energies must be the same as the stored electric potential energy it had earlier. And as we mentioned earlier, if they have equal momentum, they have equal speeds, which means they will have equal kinetic energy. So we can just say that this is simply to K kinetic energy, however, is defined as two into one half MV squared, oops and B squared. And our twos will cancel out, which means the change in potential energy for the entire system will simply just be M V squared because K one And this is solely due to the fact that K- one is equal to K- two, that's why we end up with this. So if we're trying to find the speed, right, if we're trying to find the speed, then we need to rearrange, which means we get V is equal to the square root of the change in electric potential energy over the change in mass. Now, we've already gotten our electric potential energy, we had measured it earlier. That's what we got up here. So, and we know that the mass and I'll write this in red just so we can help ourselves out. The mass of a hydrogen ion is essentially the mass of a proton, which is 1.67 times 10 to the negative 27th kilograms. So substituting all this in, we get that The speed is the square root of 1.15 times To the negative 13th jewels. Again, this answer, the change in potential energy will be the same as the work done because the work is equal to the change in potential energy and the mass In question will be 1.6, 7 times 10 to the -27 kg. Solving this gives us a final value of 8.3 times 10 to the 6th m/s. So the work done in changing the separation of the two hydrogen ions is 1.15 times 10 to the negative 13 jewels. And the speed of the separation, the speed when their separation is 1.1 times 10 to the negative 11th meters. Once it starts slipping from rest will be 8.3 times 10 to the negative or to the sixth meters per second. This aligns with answer choice D I hope this helps and I look forward to seeing you all in the next one.
Related Practice
Textbook Question
(a) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, rela­tive to infinity, is 3.75 kV? (b) What is the potential of the sphere's surface relative to infinity?
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Textbook Question
Two stationary point charges +3.00 nC and +2.00 nC are separated by a distance of 50.0 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 cm from the +3.00-nC charge?
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Textbook Question
A thin spherical shell with radius R_1 = 3.00 cm is concentric with a larger thin spherical shell with radius R_2 = 5.00 cm. Both shells are made of insulating material. The smaller shell has charge q_1 = +6.00 nC distributed uniformly over its surface, and the larger shell has charge q_2 = -9.00 nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells. (a) What is the electric potential due to the two shells at the following distance from their common center: (i) r=0; (ii) r=4.00 cm; (iii) r=6.00 cm?
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Textbook Question
A small metal sphere, carrying a net charge of q_1 = -2.80 μC, is held in a stationary position by insulat­ing supports. A second small metal sphere, with a net charge of q_2 = -7.80 μC and mass 1.50 g, is projected toward q_1. When the two spheres are 0.800 m apart, q_2, is moving toward q_1 with speed 22.0 m/s (Fig. E23.5). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of q_2 when the spheres are 0.400 m apart?
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Textbook Question
CALC. 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=q/(4πϵ_0 ) (1/r_a -1/r_b). (c) Use E_r=-∂V/∂r=(-∂/∂r) (1/(4πϵ_0 ) q/r)=[1/(4πϵ_0 )](q/r^2) and the result from part (a) to show that the electric field at any point between the spheres has magnitude E(r)=[V_ab/(1/r_a -1/r_b )](1/r^2) (d) Use E_r = [1/(4πϵ_0 )](q/r^2) 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.
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Textbook Question
CALC. 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=q/(4πϵ_0 ) (1/r_a -1/r_b). (c) Use E_r=-∂V/∂r=(-∂/∂r) (1/(4πϵ_0 ) q/r)=[1/(4πϵ_0 )](q/r^2) and the result from part (a) to show that the electric field at any point between the spheres has magnitude E(r)=[V_ab/(1/r_a -1/r_b )](1/r^2) (d) Use E_r = [1/(4πϵ_0 )](q/r^2) 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.
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