A gold(III) ion is accelerated by the electric field created between two parallel plates separated by 0.020 m. The ion carries a charge of +3q, and has a mass of 3.27 x 10 kg. A potential difference of 1000V is applied across the plates. The work done to move the ion from one plate to the other results in an increase in the kinetic energy of the gold(IIl) ion. If the ion starts from rest, calculate its final speed. 1G00. 1o-190
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- Review Multiple-Concept Example 4 to see the concepts that are pertinent here. In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 21000 V. The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen. VB =Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.32 cm. If the potential difference across the plates was 24.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.An electron moving parallel to the x axis has an initial speed of 3.7 x 106 m/s at the origin. Its speed is reduced to 1.4 x 105 m/s at x = 2 cm. Calculate the electric potential difference 1. between the origin and x -= 2 cm. 2. A proton is released from rest in a uniform electric field whose magnitude is 5000 V/m. Through what potential difference will it have passed after moving 0.25 meters? How fast will it be going after it has travelled 0.25 meters?
- An electron is to be accelerated in a uniform electric field having a strength of 2.106 (a) What energy in keV is given to the electron if it is accelerated through 0.45 m? AKE = keV m (b) Over what distance (in km) would it have to be accelerated to increase its energy by 45 GeV? d = ✔km Hint: How is potential energy, PE, gained by an electron related to the uniform electric field? How is the potential difference, V, related to the uniform electric field?Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.44 cm. If the potential difference across the plates was 29.5 kV, find the magnitude of the electric field (in V/m) in the region between the plates.In the image provide these exists two parallel plates call A and B the length of both being 5 centimeters and both separated from each other with a distance of 4 centimeters. An electron enters 1 cm above plate B and it is traveling horizontally at speed of 1x10^6 m/s. VA = 20 V and VB = 10 V are the plate potentials, respectively. A) What is the electrical field (vector) between the plates? B) What is the electric potential V at the entry point of the electron and its initial potential energy? C) The electron eventually find the speed one of the plates, find out out witch one and the speed of the electron before the impact. D) What is the location along the place where the electron makes impact (take x =0 as the left edge of the plates). OFocus 81 F 0.
- An electron is to be accelerated in a uniform electric field having a strength of 1.90 106 V/m. (a) What energy in keV is given to the electron if it is accelerated through 0.810 m?_________ keV(b) Over what distance would it have to be accelerated to increase its energy by 50.0 GeV (as in the Stanford Linear Accelerator, which is actually smaller than this)? ________kmCathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.54 cm. If the potential difference across the plates was 28.5 kV, find the magnitude of the electric field (in V/m) in the region between the plates.(a) Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 168 V. m/s (b) Calculate the speed of an electron that is accelerated through the same potential difference. m/s
- A certain parallel-plate capacitor is filled with a dielectric for which κ = 6.26. The area of each plate is 0.0573 m2, and the plates are separated by 2.35 mm. The capacitor will fail (short out and burn up) if the electric field between the plates exceeds 164 kN/C. What is the maximum energy that can be stored in the capacitor?Two plates separated by a distance d = 15.3 mm are charged to a potential difference V = 7.25 V. A constant force F = 8.31 N pushes a –8.30 mC charge from the positively charged plate to the negatively charged plate. Calculate the change in the kinetic energy AK, potential energy AU, and total energy AE of the charge as it travels from one plate to the other. Assume the initial speed of the charge is 0. AK = AU =Two charged, parallel, flat conducting surfaces are spaced d = 1.1 cm apart and produce a potential difference ΔV = 715 V between them. An electron is projected from one surface directly toward the second. What is the initial speed of the electron if its comes to rest just at the second surface?