Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens.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 1.30 cm. If the potential difference across the plates was 30.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.
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Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens.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 1.30 cm. If the potential difference across the plates was 30.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates.
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- Problem 2: Consider the parallel-plate capacitor shown in the figure. The plate separation is 4.7 mm and the the electric field inside is 26 N/C. An electron is positioned halfway between the plates and is given some initial velocity, v₁. Randomized Variables d=4.7 mm E = 26 N/C Vi,a + + + + + + + + Part (a) What speed, in meters per second, must the electron have in order to make it to the negatively charged plate? dProblem 2: Consider the parallel-plate capacitor shown in the figure. The plate separation is 5.9 mm and the the electric field inside is 24 N/C. An electron is positioned halfway between the plates and is given some initial velocity, v₁. Randomized Variables d = 5.9 mm E = 24 N/C + + + + + + + + d Otheexpertta.com Part (a) What speed, in meters per second, must the electron have in order to make it to the negatively charged plate? Part (b) If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. What will its speed be, in meters per second, when it reaches the positive plate in this case? select part Vf,b=1A small ball with charge q = 12.8 μC and mass m = 0.065 kg is suspended from the ceiling by a string of length L = 2 m and is initially at rest. A uniform horizontal electric field E of magnitude 500 V/m is applied to the ball-string system. The ball then begins to move. Ignore air resistance. a)Suppose point B is the highest point the ball can reach. Take θ as the angle of the string with the vertical direction at point B. Enter an expression for the change of gravitational potential energy ΔUg from point A to point B in terms of the symbols given. b) Enter an expression for the change of the electrical potential energy ΔUe from point A to point B in terms of the symbols given.
- A particle (m = 26 mg, q = - 5.0 µC) moves in a uniform electric field of 60 kN/C in the positive x-direction. At t = 0, the particle is moving 30 m/s in the positive x-direction and is passing through the origin. Determine the maximum distance beyond x = 0 the particle travels in the positive x-direction. Select one: A. 19.5 m B. 117.0 m C. 78.0 m D. 39.0 mConsider horizontal parallel plates with a fix potential difference. The upper plate has a voltage difference of 30 V with a lower plate, and they are separated by 3 cm. You move the two plates carefully to a separation of 4 cm. What is the strength of the electric field between the plates? Answers 750 volts/meter. Would like to see how to get that answer.An electron is to be accelerated in a uniform electric field having a strength of 2.3·106 Vm. a) What energy in keV is given to the electron if it is accelerated through 0.35 m? b) Over what distance (in km) would it have to be accelerated to increase its energy by 58 GeV?
- An alpha particle is a doubly ionized helium nucleus (which means it has a +2e charge). You are asked to accelerate an alpha particle within a uniform electric field of 1.6 MV/m (here the "M" stands for Mega, which means millions), for a distance of 0.02 meters. What would be the kinetic energy of the alpha particle at the end of this distance? Please enter your answer in keV (kilo electron volts). This means if you find 3500 eV as your answer you need to enter 3.5 to the box below.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 1.56 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. ????V/mA proton (m=1.67 x 10^-27 kg) travels a distance of 4.3 cm parallel to a uniform electric field 3.5 x 10^5 V/m between the plates shown in the figure. If the initial velocity is 3.2 x 10^5 m/s, find the magnitude of its final velocity in m/s. (Ignore gravity)
- 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?A doubly charged ion is accelerated from rest to a kinetic energy of 31.5 keV by the electric field between two parallel conducting plates separated by 1.75 cm. What is the electric field strength between the plates, in volts per meter?