PHYSICS F/SCI.+ENGR.,CHAPTERS 1-37
5th Edition
ISBN: 9780134378060
Author: GIANCOLI
Publisher: RENT PEARS
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When an electron is accelerated from
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Suppose an electron (q= -e = -1.6 x 10^-19 C, m = 9.1 x 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. Use the template in the attached photo to solve for the problem.
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Chapter 23 Solutions
PHYSICS F/SCI.+ENGR.,CHAPTERS 1-37
Ch. 23.1 - Prob. 1AECh. 23.2 - Prob. 1BECh. 23.3 - Prob. 1CECh. 23.3 - Prob. 1DECh. 23.8 - Prob. 1FECh. 23.8 - Prob. 1GECh. 23 - If two points are at the same potential, does this...Ch. 23 - If a negative charge is initially at rest in an...Ch. 23 - State clearly the difference (a) between electric...Ch. 23 - Suppose the charged ring of Example 238 was not...
Ch. 23 - Consider a metal conductor in the shape of a...Ch. 23 - Equipotential lines are spaced 1.00 V apart. Does...Ch. 23 - Prob. 1PCh. 23 - Prob. 3PCh. 23 - Prob. 4PCh. 23 - Prob. 9PCh. 23 - Prob. 11PCh. 23 - (II) The electric potential of a very large...Ch. 23 - (II) The Earth produces an inwardly directed...Ch. 23 - (II) A 32-cm-diameter conducting sphere is charged...Ch. 23 - (II) An insulated spherical conductor of radius r1...Ch. 23 - (II) Determine the difference in potential between...Ch. 23 - (II) Suppose the end of your finger is charged....Ch. 23 - (II) Estimate the electric field in the membrane...Ch. 23 - (III) A hollow spherical conductor, carrying a net...Ch. 23 - (III) A very long conducting cylinder (length ) of...Ch. 23 - Prob. 31PCh. 23 - (I) Draw a conductor in the shape of a football....Ch. 23 - (II) Equipotential surfaces are to be drawn 100 V...Ch. 23 - (II) Calculate the electric potential due to a...Ch. 23 - (III) The dipole moment, considered as a vector,...Ch. 23 - (I) Show that the electric field of a single point...Ch. 23 - (I) What is the potential gradient just outside...Ch. 23 - (II) The electric potential between two parallel...Ch. 23 - () The electric potential in a region of space...Ch. 23 - (II) In a certain region of space, the electric...Ch. 23 - (II) A dust particle with mass of 0.050 g and a...Ch. 23 - (III) Use the results or Problems 38 and 39 to...Ch. 23 - (I) How much work must be done to bring three...Ch. 23 - (I) What potential difference is needed to give a...Ch. 23 - If the electrons in a single raindrop, 3.5 mm in...Ch. 23 - By rubbing a nonconducting material, a charge of...
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- Suppose 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_forwardA uniform electric field is created by two horizontal parallel plate electrodes of length 10 cm separated by a vertical distance d = 5 cm and held at a potential difference V = 250V. A beam of electrons is introduced between the two plates with a horizontal velocity v = 5 x 106 m/s. What is the deflection distance towards the positive plate when the electron beam leaves the electric field?arrow_forwardAnswers: a) 3.3 C c/m² 6. Two electrons are fixed 2.00 cm apart. Another electron is shot from infinity and comes to rest midway between the two. What was its initial speed? Answer: 318 m/sarrow_forward
- If electrons have kinetic energy of 2000 eV, find (a) their speed, (b) the time needed to traverse a distance of 5 cm between plates that our horizontal, and (c) the vertical component of their velocity after passing between the plates if the electric field is 3.33 x 10^3 V/m.arrow_forwardA small charged ball is accelerated from rest to a speed v by a 500 V potential difference. If the potential difference is changed to 16,000 V, what will the new speed of the ball be? O 16v 0 4v O 2v W.arrow_forwardA small charged particle of mass 1.0 x 10-8 kg is traveling rightward between two plates separated by a distance d = 80 cm, as shown below. The electric field between the plates has a constant magnitude of 3.0 x 106 V/m and is directed leftward. The particle's speed is 5.0 x 103 m/s at the left plate and 2.0 x 10³ m/s at the right plate. Ignore the effect of gravity. F (a) Is the particle positively charged or negatively charged? Justify your answer briefly but clearly. (b) Find the charge (with correct sign) of the particle, as well as the potential difference (with correct sign) through which the particle has moved. (Note: The potential difference is positive if the right plate is at a higher potential than the left plate, and negative if the right plate is at a lower potential than the left plate. Show all your work; do not simply plug numbers into a result derived in class.)arrow_forward
- This scanning electron microscope image of a bacterium was produced using a beam of electrons accelerated through an 30 kV potential difference. What is the speed of the electrons?arrow_forwardA positron (a particle with a charge +e and a mass equal to that of electron) that is accelerated from rest between two points at a fixed potential difference acquires a speed of 9.0x10^7 m/s. What speed is achieved by a proton accelerated from rest between the same two points? (Disregard relativistic effects.) a) 2.5x10^6 m/s b) 2.1x10^6 m/s c) 2.8x10^7 m/s d) 4.9x10^7m/s e) None of the Abovearrow_forwardFour parallel plates are connected in a vacuum as shown in the picture. An electron with initial velocity, 1.02 x 10°m/s in the hole of plate X is accelerated to the right. Gravity is negligible once the electron passes through holes at W and Y. However due to the high air viscosity, the electrons loses 1.6 × 10-17) of 1V = energy between the plate W to plate Y. It then passes through the hole at Y and slows down as it heads to plate Z. Calculate the distance, in centimetres, from plate Z to the point at which the electron changes direction. - 6.0 cm→6.0 cm→<6.0 cm - W Y 4.0 x 10² V 7.0 × 103 V Narrow_forward
- What is the speed of an electron after being accelerated from rest through a 2.5×107 V potential difference? Express your answer as a fraction of cc.arrow_forwardThe poles of a 15ov battery are connected to two parallel plates with an area of 30 cm? and a distance of 10 mm between them, as shown in the figure. A beam of alpha particles (9.=+2lqel, Iqel=1.6x1019 C and m=6.64x1027 kg) 2 - starts from rest, accelerates with a potential difference of 10 kV and enters the interplate region with an electric field perpendicular to the direction of motion. What is the magnitude and direction of the magnetic field needed for alpha particles to emerge from between the plates without deviating? .150 V B=0.0612 T outward from the plane of the page b) B=0.0306 T outward from the plane of the page B=0.0153 T inward from the plane of the page B=0.0153 T outward from the plane of the page c) d) B=0.0306 T inward from the plane of the pagearrow_forwardThe uniform electric field between two charged parallel plates has a magnitude of 4400 V/m (note that 1 V/m = 1 N/C). If the plates are separated by 0.0100 m and an electron is released from the far negative side of the gap, what will the electron's kinetic energy (in eV) and speed be when it just reaches the positive plate? Note that the electron charge = -1.602 x 10-19 C and the electron mass = 9.11 x 10-31 kg. KE V= ev m/sarrow_forward
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