A very small electron gun accelerates electrons, starting from OV rest, through an electric potential difference of AV, = 140 V inside the gun. Its end is inside a capacitor, oriented at an 1 max angle 0. = 45%, the capacitor plates are square with a side of 10.0 cm and are spaced a distance d = 4.00 mm apart. The AV electrons emerge from the gun at an initial height of y. = 2.50 Electron gun mm as shown, reach a maximum height ymax as shown, and continue following a parabolic path until hitting the positive plate, which is at a potential V. = +200 V relative to the other plate. (Note: The potentials inside the electron gun and inside the capacitor are unrelated) a) Find the speed v. of an electron when it emerges from the electron gun, then write its velocity as a vector v. in component form. b) Find the electric field strength E inside the capacitor and draw its vector on the diagram. c) Use this to find the surface charge density n and the total charge O on the capacitor plates. d) Find the value of ymax. How close does the electron come to the top plate? i) Write the electron's velocity as a vector v, when it reaches y max' e) Find the electron's velocity vector v2 when it hits the positive plate. Note: you can either just work with the y component of velocity alone, ignoring the x component, or work with the total velocity and find the y component from that. i) Find the angle 0 at which it hits the plate, as shown on the diagram. f) (Small extra credit): Your calculation is not exactly right because you did not account for polarization of the capacitor plates when the electron is near them. A charge produces its own E field, which must everywhere meet the conductor at a right angle. If you think of the surface of the conductor as if it were a mirror so the electron could see an image of itself behind the mirror, except it was positive like a proton, the field lines going from the electron to the surface and continuing to the imaginary proton would be exactly the same as if you replaced the mirror image with a real proton at the same position. Therefore you can calculate the force on the electron by pretending there is an equal but opposite charge at an equal distance beyond the conductor surface, and find its field. This is a powerful method called the method of images. Note that there will be an image charge created in both capacitor plates, one above the top plate and one below the bottom plate, but each one twice as far away as the distance from the electron to that plate. Make a drawing and calculate the field strength of the image charge at the location of the electron when it is at ymay Comment on how this image field compares to the electric field strength of the capacitor. Predict what effect this real image field will have on the path of the electron, including which plate it hits and where
