An electron is to be accelerated in a uniform electric field having a strength of 2x10^6 V/m. Calculate: a) What energy in kev is given to the electron if it is accelerated through 0.4 m?
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- A parallel-plate capacitor has a plate separation of 1.5 mm and is charged to 900 V. 1) If an electron leaves the negative plate, starting from rest, how fast is it going when it hits the positive plate? (Express your answer to two significant figures.) ×10¹ km/sSuppose 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=Considering electron and proton as two charged particles separated by d = 4.3 x 10-11 m calculate the electrostatic potential at a distance d from the proton. Take the mass of the proton 1.7 x 10-27 kg, the mass of the electron 9.1 x 10-31 kg, the electron charge -1.6 × 10-¹⁹ 1 C and = 9 × 10⁹ m/F. Απευ Answer: Choose...
- The electric field strength between two parallel conducting plates separated by 3.60 cm is 4.90 ✕ 104 V/m. a) What is the potential difference between the plates (in kV)? and b)The plate with the lowest potential is taken to be at zero volts. What is the potential (in V) 1.00 cm from that plate (and 2.60 cm from the other)?The difference in electric potential between the accelerating plates in the electron gun of a TV tube is 25 kV, while the distance between them is 1.5 cm. The magnitude of the uniform electric field between the plates is 3.8 10 2 v/m. 5.6 10² v/m. O 1.7 106v/m. 11'108 v/m.(a) Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 144 V. 166.11*10^3 (b) Calculate the speed of an electron that is accelerated through the same potential difference. 7.12*10^6
- A 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 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?
- A Proton Gun. You and your team are designing a device that creates a beam of high-velocity protons. The device consists of a parallelplate capacitor in which protons start at rest at the positive plate. The protons accelerate toward the negative plate, which has a hole through which some of the protons pass. The plates are separated by 5.6cm, and the potential difference between the plates is 5.2kv. (a) What are the kinetic energies of the protons when they pass through the hole? (b) What are the speeds of the protons when they pass through the hole? (c) What is the acceleration of the protons between the plates? (a) Number UnitsAn electron initially at rest, is allowed to accelerate through a potential difference of 1 V, gaining kinetic energy KEe, whereas a proton, also initially at rest, is let accelerate through a potential difference of 1 V, gaining kinetic energy KEp. As ∣qe∣ = ∣qp∣ but mp >> me, therefore, Group of answer choices KEe >> KEp KEe << KEp KEe = KEp All we can say, is KEe ≠ KEpCathode 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.