An electron with a horizontal speed of 4600000 m/s and no vertical component of velocity passes through two horizontal parallel plates, as shown. The magnitude of the electric field between the plates is 240 N/C. The plates are 6 cm long. Calculate the angle to the horizontal, 0 (in degrees), that the electron would pass through the plates with. e + + + + + + YY y X
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- Consider an electron that is 1010 m from an alpha particle (q=3.21019C) . (a) What is the electric field due to the alpha particle at the location of the electron? (b) What is the electric field due to the electron at the location of the alpha particle? (c) What is the electric force on the alpha particle? On the electron?An electron with a horizontal speed of 4200000 m/s and no vertical component of velocity passes through two horizontal parallel plates, as shown. The magnitude of the electric field between the plates is 150 N/C. The plates are 3 cm long. Calculate the angle to the horizontal, 8 (in degrees), that the electron would pass through the plates with. + 0 y -XTwo large parallel plates are placed horizontally with respect to each other and are then connected to a100-V battery. If the distance between the plates is 1 cm, the uniform electric field between the plates is104 N/C in the upward direction as shown by the vectors in the figure. If the electron is released from restat the upper plate, calculate:i. kinetic energy of the electron after traveling 1 cm to the lower plate, andii. the time required for it to travel this distance.
- Here’s one more model of an electron. Here e is the magnitude of the charge of an electron, and R is a constant parameter characterizing the electron’s size: rho=-e (105/4pi)r2(R-r)2/R7 , rho = 0 for r >R. a. Use the problem's symmetry to evaluate the electric field at all points in space, it r >R and r<R. b. Sketch the electric field magnitude versus r for all values of r. d. evaluate the electric potential at all points in space and sketch its magnitude as a function of r.Ex : A particle ( mass = 5.0g, charge = 40mC ) move in a region of space where the electric field is uniform and is given by E, = 2.5N/c,Ey = E, = 0. If the velocity of particle at t 0 is given by vy = %3! 50m/s,v = v,y = 0, what is the speed of particle at t= 2. Os.An electron flies into a constant electric field (along the direction of the lines of E-field). The initial electron velocity is 10 km/s. Calculate the magnitude the electric field if the electron stops in 6 nsec. The electron mass is me =9.11×10-31 kg, the electron charge is qe = -1.61×10-19 C. The electric field, E = Units .
- Question 1 a) In J. J. Thomson experiment (1897), an electron moving horizontally with a constant speed vo enters in between the horizontal plates of a capacitor. The electric field strength between the plates of length L and distance d, is E. The vertical deviation of the electron at the moment of exit from the field region is measured to be Y. Derive the expression giving the electron's charge to mass ratio, i.e. e/m to be 2v,Y/CEL). (Recall that Thomson received Nobel Prize for his achievement.) b) Calculate e/m, knowing the following data. E=1.6x10* Newton/Coulomb, L=10 cm, Y=2.9 cm, v=2.19x10* km/s. (Be careful to use coherent units.)An electron with kinetic energy K is traveling along the positive x-axis, which is along the axis of a cathode‑ray tube, as shown in the figure. There is an electric field E=18.0×104 N/C pointed in the positive y-direction between the deflection plates, which are 0.0600 m long and are separated by 0.0200 m. Determine the minimum kinetic energy ?minKmin the electron can have and still avoid colliding with one of the plates.A thundercloud produces a vertical electric field at a magnitude of 2.8 x10^4 N/C at ground level. You hold a 22.0 cm x 28.0 cm sheet of paper horizontally below the cloud. What is the electric flux through the sheet? An electric field of 500 V/m is destred between two parallel plates 153.0 mm apart. What voltoge should be applied?
- Find the electric field (magnitude and direction) at the location of qa in the figure below, given that qp = 10.70 µC and qc = -5.90 µC. Point charges located at the corners of an equilateral triangle 22.0 cm on a side. qa Magnitude |E| direction (wrt +x-axis) Submit Answer Incompatible units. No conversion found between "uc" and the required units. O Tries 0/10 Previous Tries What is the force (magnitude and direction) on qa, given that qa = 1.40 nC. Magnitude |F| direction (wrt +x-axis) Submit Answer Tries 0/10An electron moves in uniform circular motion in a circle of radius r equal to 0.1 cm around an infinitely long wire with a linear charge density ∂ of 0.14 µC/m a. Use gauss law to find the magnitude of the electric field at the distance r from the wire. b. The mass of the electron is 9.11x10^-31 kg and its electric charge equals 1.6x10^-19 C. Calculate the period of the revolution of the uniform circular motion.electrons Two protons (p) and two (e) are arranged on a circle of 5 [cm], with angles 0₁ = 20°, 0₂ = 60°, 03 = 20° and 04 = 60°, as radius r shown in the figure. (qp = +1.6 x 10-1⁹ [C] and qe = -1.6 × 10-19 [C]). The figure is not to scale. 04 y a. Find Enet, the net electric field vector produced at the center of the circle. Enet =([ ])i + ( ]) Ĵ [N/C] b. Where on the circle should a fifth point charge qo be placed (give its angle relative the +x-axis) and what is its value (calculate qo) in order to have Ēnet (the net electric field at the center of the circle) equals zero (Type the detailed solution to this question in the below box, Show all your calculation steps by typing in the box). Р X