Find the electric field (in units of N/C) needed to accelerate a charged particle (m-22 ug and q=33 u) from rest to a speed of 100m/s in a distance of 575.4 cm. Select one: OA.403.76 OB.228.21 OC 579.31 OD 263.32 OE 754.85
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Q: Question uploaded with figure...thanks
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- An electron moving through an electric field experiences an acceleration of 4.3 ✕ 103 m/s2. (a) Find the electric force acting on the electron. N in the same or opposite direction as E.(b) What is the strength of the electric field? N/CA constant electric field accelerates a proton from rest through a distance of 1.90 m to a speed of 1.57 × 10^5 m/s. ( The mass and charge of a proton are mp= 1.67×10^-27 kg ans qp= e= 1.60 × 10^-19 C.) a) Find the change in the protons kinetic energy ( in J). b) Find the change in the system's electric potential energy ( in J). c)Calculate the magnitude of the electric field ( in N/C).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 evacuated tube uses an accelerating voltage of 30 kV to accelerate electrons to hit a copper plate and produce X-rays. Non-relativistically, what would be the maximum speed (in m/s) of these electrons? 3.25E-07 × m/ss In the microscopic view of electrical conduction in a copper wire, electrons are accelerated by an electricfield and then collide with metal atoms after traveling about3.9 * 10-8m. If an electron begins from rest and is accelerated bya field of 0.065 N>C, what is its speed when it collides with a metalatom?An electron (m 9.11x10 31 kg) accelerates in a uniform electric field E = 1.45 x 10 N/C. What is the acceleration of the electron after it traveled 10 cm in the electric field? O A. 2.55 x 1014 m/s² O B. 2.55 x 1015 m/s? C. 1.5 x 1014 m/s? O D. 1.5 x 1015 m/s² O E. None Nex to search
- b) The uniform electric field between two parallel and opposite charge plates separated by 30 mm has an intensity of 2.5 x10³ N/C. An electron is released from the negative plate with zero initial velocity. What is the kinetic energy of the electron at the halfway point.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?An electron and a proton, each starting from rest, are accelerated by the same uniform electric field of 260N/C. Determine the distance (d) and time (t) for each particle to acquire a kinetic energy of 5.0x10^-17 J
- helpQ2. A proton at speed v =3.00 × 10ʻ m/s orbits at radius r = 1.00 cm outside a charged sphere. Find the sphere's charge.Two large aluminum plates are separated by a distance of 2.0 cm and are held at a potential difference of 195 V. An electron enters the region between them at a speed of 3.2 × 105 m/s by passing through a small hole in the negative plate and continues moving toward the positive plate. Assume the electric field between the plates is uniform. 1. What is the electron’s speed, in meters per second, when it is 0.1 cm from the negative plate? 2. What is the electron’s speed, in meters per second, when it is 0.5 cm from the negative plate? 3. What is the electron’s speed, in meters per second, when it is 1.0 cm from the negative plate? 4. What is the electron’s speed, in meters per second, when it is 1.5 cm from the negative plate? 5. What is the electron’s speed immediately before it strikes the positive plate?