Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 21, Problem 54P
To determine
The speed of the electron.
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In large CRT televisions, electrons are accelerated from rest by a potential difference of 23.88 kV
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Answer:
x10' m (express your answers in tenths place or one decimal digit only)
An electron is located at the origin of the coordinates, anda second electron is brought to a position 2 Å from theorigin in the 1y direction.(a) Calculate the force on the electron at the origin, givingthe components Fx and Fy.(b) Calculate the potential energy of the two electrons.
An electron is accelerated from rest through a potential difference of 3.00 kV. What is its final velocity? The mass of an electron is 9.109×10-31 kg.
Chapter 21 Solutions
Physics for Scientists and Engineers
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- ... Path of trajectory AV w An electron is fired at a speed v¡ = 3.4 x 106 m/s and at an angle 0; = 30.5° between two parallel conducting plates as shown in the figure. If s = 1.5 mm and the voltage difference between the plates is AV = 98 V, determine how close, w, the electron will get to the %3D %3D bottom plate. Put your answer in meters and include at 6 decimal places in your answer. Do not include units. The x-axis of the coordinate system is in the middle of the parallel plate capacitor. Round your answer to 6 decimal places.arrow_forwardSuppose an electron (q = -e = -1.6 x 10¬9 C,m=9.1 x 10¬3' kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for %3D the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U = 0 K = -U 1 Since K mv and using the formula for potential energy above, we arrive at an equation for speed: 2 v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=arrow_forwardThe friends now try a homework problem. Consider an electron, of charge magnitude e = 1.602 × 10-¹⁹ C and mass me = 9.11 x 10-31 kg, moving in an elect field with an electric field magnitude E = 7 x 10² N/C, similar to what Thana observed in the simulation. Let the leng the plates be L = 50 cm, and the distance between them be d = 20 cm. Find the maximum speed, v, the electron co be moving if it enters the space halfway between and parallel to the two plates to just barely strike one of the plates 7.01e+06 X m/s If the field is pointing upward, which plate will Thana conclude the electron strikes at this speed? The lower plate, because the electron is negatively charged. The lower plate, because the electron is attracted to the negative plate. The upper plate, because we are only considering the magnitude of the electron charge, and magnitudes are always positive. O The upper plate, because the electron charge magnitude is positive.arrow_forward
- Suppose an electron (q = - e = - 1.6 x 10¬19 C,m=9.1 × 10¬3' kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K + U = 0 K = -U mv and using the formula for potential energy above, we arrive at an equation for speed: 2 Since K= v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=arrow_forwardSuppose an electron (q= -e= -1.6 x 10-19 C,m=9.1x 10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K +U = 0 K = -U Since K- and using the formula for potential energy above, we arrive at an equation for speed: v = ( 51/2 Plugging in values, the value of the electron's speed is: V= x 107 m/sarrow_forwardInitially at rest, an electron is accelerating toward a target that is 20,000V higherpotential than where the electron started. What is the speed of the electron just as itreaches the target? Note that the charge of an electron is -1.6x10-19 C and its mass is9.1x10-31 kg. Imagine that we had two electrons that were initially VERY far apart, moving towardeach other at nearly the speed of light, roughly 3 x 108 m/s. How close can they get toeach other (after all, like charges repel!)? A 0.0012 C charged cart is initially at rest on a ramp (1.1 m away from the bottom of theramp). At the bottom of the ramp is another charge (0.012 C). Ignoring friction, how fastis this cart (assume 100 kg) moving after it has gone 20 m up the ramp? Assume that theramp is 30 deg. from the horizontal.arrow_forward
- (a) Find the potential difference ΔⅤe required to stop an electron (called a "stopping potential") moving with an initial speed of 2.12 x 107 m/s. .128 Your response is off by a multiple of ten. kVarrow_forwardin a straight line with a 6° Omis. I t 'enters underges migh along the same straight line, An electron is moving a region 5.1 m Olong where it on ocelerotion of where it on occelerotion 5.1um 12 6.2.10" What is the electrons velo city in (mis) when i this emerges from tis region? mis bl How lang regien? Cins) does the electron take to cross thearrow_forwardHow much work is needed to assemblean atomic nucleus containing three protons (such as Li) if we modelit as an equilateral triangle of side 2.00 * 10^-15 m with a proton at eachvertex? Assume the protons started from very far away.arrow_forward
- In nuclear fission, a nucleus splits roughly in half. What is the potential 4.00 x 10-14 m from a fragment that has 50 protons in it? (p=+1.602x10-19 C)arrow_forward(27) D3.6. Find the energy stored in free space for the region 2 mm < r < 3 200 mm, 0 < 0 < 90°, 0 < < 90°, given the potential field V =: (a) = - V: r 300 cos @ (b) V. Ans. 46.4 J: 36.7 Jarrow_forwardSuppose an electron (q= -e= -1.6x 10-19 C.m=9.1×10-31 kg) is accelerated from rest through a potential difference of Vab = +5000 v. Solve for the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U- Assuming all potential energy U is converted to kinetic energy K. K+U=0 K--U Since K=mv and using the formula for potential energy above, we arrive at an equation for speed: 1/2 Plugging in values, the value of the electron's speed is: x 10' m/sarrow_forward
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