Physics 21 Fall, 2014 Solution to HW-6 23-8 Three equal 1.05 µC point charges are placed at the corners of an equilateral triangle whose sides are 0.600 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.) q1 0.6m 0.6m q2 0.6m q3 To do this problem, we need to know that the electric potential energy for any two charges qi and qj separated by a distance rij is Uij = 1 qi qj . 4πǫ0 rij When we have several charges, we invoke superposition and just add up the contributions from each distinct pair: 1 X qi qj . U= 4πǫ0 i<j rij In our specific situation, using q1 , q2 , and q3 there are three distinct pairs, and the equation looks like this: 1 q2 q3 q1 q3 q1 q2 U= . + + 4πǫ0 r12 r23 r13 For this problem, r12 = r23 = r13 = 0.6 m, and q1 = q2 = q3 = 1.05 × 10−6 C. Since all the charges are equal, and all the distances are equal, we get the total U by multiplying the potential energy of any one pair by three: 2 1.05 × 10−6 C U = 3 9 × 109 Nm2 /C2 0.6 m = 0.0496 J We can use conservation of energy to help solve this problem. After the protons have been accelerated by the cyclotron, we can assume they are isolated and only interact with each other. Their energy will be a combination of kinetic and electrical potential energy. Energy is conserved in this isolated system, and we can write the total energy as Etotal = KE + P E = 2 12 mp v02 + U0 = mp v02 + U0 where mp is the mass of the proton, v0 = 1.6 × 106 m/s is the initial velocity of each proton, and U0 is the initial electrical potential energy between the two protons. The electrical potential energy between two protons (which 1 e2 /r, and r is the distance between have charge e) is 4πǫ 0 the protons. As the protons approach one another (part A in figure), r decreases and so the potential energy increases. This results in a decrease in the kinetic energy of the protons to keep the total energy constant. Eventually, all the kinetic energy of the protons has been converted to electrical potential energy at some minimum distance of approach (part B), and the protons will momentarily be at rest. They will then begin to separate due to their repulsion and the potential energy will be converted back to kinetic energy (part C). 1 The force between the two protons is 4πǫ e2 /r2 , and so to 0 find the maximum force, we must find the minimum distance of approach. At this minimum distance, rmin , the total energy is all potential energy since the protons are at rest. Using conservation of energy, mp v02 + U0 = 1 e2 4πǫ0 rmin ⇒ rmin = 1 e2 4πǫ0 mp v02 + U0 No information was given on the initial separation of the protons, so assume they were far apart to begin with and 1 then U0 = 4πǫ e2 /r0 ≈ 0 for a large value of r0 . Then 0 rmin 2 9 × 109 Nm2 /C2 1.602×10−19 C e2 1 = = 2 4πǫ0 mp v02 1.67 × 10−27 kg (1.6 × 106 m/s) = 5.40 × 10−14 m We can either substitute the algebraic expression or the numerical value of 1/rmin into the Coulomb force law. The former method gives m2p v04 1 e2 = 4πǫ 0 2 4πǫ0 rmin e2 −27 (1.6726 × 10 kg)2 (1.6 × 106 m/s)4 = (8.99 × 109 Nm2 /C2 )(1.602 × 10−19 C)2 Fmax = = 7.97 × 10−2 N. 23.12 Two protons are aimed directly toward each other by a cyclotron accelerator with speeds of 1600 km/s, measured relative to the earth. Find the maximum electrical force that these protons will exert on each other. This force may be miniscule by our standards, but for a proton it corresponds to an acceleration of 4.77 × 1025 m/s2 ! September 14, 2014 23-13 A small particle has charge −5.10 µC and mass 2.30 × 10−4 kg. It moves from point A, where the electric potential is VA = 300 V , to point B, where the electric potential VB = 690 V is greater than the potential at point A. The electric force is the only force acting on the particle. The particle has a speed of 3.10 m/s at point A. (a) What is its speed at point B ? (b) Is it moving faster or slower at B than at A? (a) This problem is easily solved using the conservation of energy. Knowing the change in electric potential and the charge on the particle allows us to calculate the change in electric potential energy. Then the change in kinetic energy will allow us to calculate the final speed of the particle. UA + K A = UB + K B 2 2 qVA + 21 mvA = qVB + 12 mvB 2q 2 2 vB = vA + (VA − VB ) m vB = s 2 (3.10 m/s) + = 5.19 m/s 2 (−5.10 µC) (300 V − 690 V) 2.30 × 10−4 kg Don’t forget to enter ×10−6 for µ when you type the charge into your calculator. Note that even though the particle is gaining electric potential, it is losing electric potential energy. This is due to the negative charge on the particle. Thus, it is gaining kinetic energy, and therefore, speed. (b) Faster electron’s charge to get the electrical potential energy of the electron in this potential field. U (x) = − Qe 1 √ 2 4πǫ0 x + a2 We apply conservation of energy, U0 + K 0 = U1 + K 1 , where U0 and K0 are the PE and KE at the initial position x0 = 0.320 m, and U1 and K1 are the PE and KE at the final position x1 = 0. Knowing that the initial kinetic energy is zero, we get K1 = U0 −U1 , where K1 = 12 me v12 , U0 = U (x0 = 0.320 m), U1 = U (x1 = 0), me is the electron mass, and v1 is the speed of the electron when it reaches the center of the ring. Solving for v1 and plugging in the values gives us our answer: v ! u r u 1 2Qe 1 2 1 t v1 = (U0 − U1 ) = −p me 4πǫ0 me a x20 + a2 s N · m2 2(20.0 nC) (1.60 × 10−19 C) = 8.99 × 109 C2 9.11 × 10−31 kg v u 1 1 ×u −q t 0.140 m 2 2 (0.320 m) + (0.140 m) = 1.65 × 107 m/s This is not part of the question, but it is useful to think about what happens to the electron after it passes through the ring. The electron’s potential energy as a function of position along the axis is shown in the plot below. Electron Potential Energy -6 We need to solve this problem using conservation of energy. The electron starts from rest, but it feels the attractive force of the positive charges, and starts to move along the axis toward the ring. As it moves, it is gaining kinetic energy and losing electrical potential energy. We can calculate the electrical potential energy of the electron using the formula U = qV , where q is the electron’s charge and V is the electric potential of the ring of charge. We already know the formula for electric potential along the axis of a ring of charge from lecture and from Example 23-11 in the textbook. V (x) = Q 1 √ 2 4πǫ0 x + a2 where Q is the total charge on the ring, x is the distance along the axis, and a is the radius of the ring. Note that this equation is not on the equation sheet. We multiply by the -8 Energy (10-17 J) 23-29 A uniformly charged thin ring has radius 14.0 cm and total charge 20.0 nC. An electron is placed on the ring’s axis a distance 32.0 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. Find the speed of the electron when it reaches the center of the ring. -10 -12 -14 -16 -18 -20 -22 -0.4 -0.3 -0.2 -0.1 0 0.1 0.2 0.3 0.4 Position along axis (m) It can be seen that the minimum potential energy occurs at the center of the ring. This must be where the electron has the most kinetic energy. Once the electron passes through the ring, it will begin to gain potential energy, and lose kinetic energy. Because the potential energy is symmetric about the ring, the electron will come to rest behind the ring at the same distance along the axis from which it was released. It will then turn around and pass through the ring again. It will continue to oscillate in this manner. 23-68 A disk with radius R has a uniform charge density σ. (a) By regarding the disk as a series of thin concentric rings, calculate the electric potential V at a point on the disk’s axis a distance x from the center of the disk. Assume that the potential is zero at infinity. (Hint: Use the result that the potential at a point on the ring axis at a distance x from the center of the ring is V = 1 Q √ 2 4πǫ0 x + a2 where Q is the charge of the ring.) (b) Find Ex = −∂V /∂x. (a) Using the hint, we can modify the result for the finite ring to determine the potential dV due to a thin ring of radius r and charge dQ. We have dV = dQ 1 √ . 4πǫ0 x2 + r2 Because electric potential is a scalar quantity, we can integrate dV without keeping track of vectors. First, we must determine dQ in terms of known quantities. It is equal to the surface charge density times the surface area dA of the thin ring, dQ = σdA = σ2πr dr, where dA is the product of the length (circumference) 2πr and width dr of the ring of radius r. Now it is possible to do the integration Z R Z R Z σ 2πσr dr r dr 1 √ √ = V = dV = 2 2 4πǫ0 0 2ǫ0 0 x +r x2 + r2 The integral is on the equation sheet: Z p u du √ = a2 + u 2 . a2 + u 2 Evaluating it at the limits results in i σ hp 2 V = x + R2 − x . 2ǫ0 (b) The x component of the electric field is equal to −∂V /∂x: 1 σ 1 2 ∂V 2 −2 x +R =− − 2x − 1 ∂x 2ǫ0 2 σ x . = 1− √ 2 2ǫ0 x + R2 Notice that as R → ∞, this result reduces to the field of an infinite sheet.

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