What voltage will accelerate protons to a speed of 6*10^6 m/s?
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What voltage will accelerate protons to a speed of 6*10^6 m/s?
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- A proton is located at the origin, and a second proton is located on the x-axis at x1 = 6.22 fm (1 fm = 10−15 m). (a) Calculate the electric potential energy associated with this configuration. J (b) An alpha particle (charge = 2e, mass = 6.64 ✕ 10−27 kg) is now placed at (x2, y2) = (3.11, 3.11) fm. Calculate the electric potential energy associated with this configuration. J (c) Starting with the three particle system, find the change in electric potential energy if the alpha particle is allowed to escape to infinity while the two protons remain fixed in place. (Throughout, neglect any radiation effects.) J (d) Use conservation of energy to calculate the speed of the alpha particle at infinity. m/s (e) If the two protons are released from rest and the alpha particle remains fixed, calculate the speed of the protons at infinity. m/sAn electron at position A at 2.5×10–11m form the nucleus of an atom as shown in the picture if the electron moves to position B at 4.2×10–12m from the nucleus. What is the potential difference during the movement of the electron from point A to point B? (e= 1.6×10–19C) A)28V B)-34.5V C)-30V D)-23.4vConsider a point charge q = 0.93 µC, point A at distance d₁ = 1.7 m from q, and point B at distance d₂ = 1.0 m. (a) If A and B are diametrically opposite each other, as in Figure (a), what is the electric potential difference VA - VB? (b) What is that electric potential difference if A and B are located as in Figure (b)? (a) Number i (b) Number i B 9 (a) Units Units A d (b) A
- you have a nucleus with 19 protons, placed at x = 3.6 Angstroms. What what TWO values of x (along the x-axis) will the total electrostatic potential V be equal to zero?In the figure a charged particle (either an electron or a proton) is moving rightward between two parallel charged plates separated by distance d = 6.10 mm. The plate potentials are V₁ = -67.0 V and V₂ = -50.0 V. The particle is slowing from an initial speed of 85.0 km/s at the left plate. (a) Is the particle an electron or a proton? (b) What is its speed just as it reaches plate 2?Consider a point charge q = 0.81 µC, point A at distance d₁ = 1.9 m from q, and point B at distance d₂ = 0.93 m. (a) If A and B are diametrically opposite each other, as in Figure (a), what is the electric potential difference VA - VB? (b) What is that electric potential difference if A and B are located as in Figure (b)? (a) Number (b) Number i B (a) Units Units < d₁ (b) A
- A proton enters a parallel plate system moving with an initial velocity of 9413.1 m/s perpendicular to the electric field. The proton enters at the the positive plate and the electric field between the plates moves it in the direction of the negative plate. The plate separation distance is 19.2 cm and the potential difference between the plates is 1.9 V. What is the magnitude of the final velocity of the proton when it hits the negative plate? ( Please state what kinematics formula was used to find final velocity through the steps.)Now you have a nucleus with 16 protons at x = 2.7 Angstroms on the x-axis. What is the value of the electrostatic potential V at a point on the positive y-axis, at y = 3.1 Angstroms?Consider a point charge q = 0.87 μC, point A at distance d₁ = 2.5 m from q, and point B at distance d2 = 1.4 m. (a) If A and B are diametrically opposite each other, as in Figure (a), what is the electric potential difference VA - VB? (b) What is that electric potential difference if A and B are located as in Figure (b)? (a) Number M. i (b) Number i B -d₂4 9 (a) Units Units -d- A B de dj- (b) 4