Question

Name PROBLEM 3-8 points An electron enters an empty parallel plate capacitor that is charged to a potential difference of 1000 volts. The capacitor has a gap of 1 mm in the y direction between the plates [ po for each graph] On each graph, plot the indicated quantity, as y varies fromy -0 mm to y- 1 mm, across the gap (at the center of the plates). Show the shape of each curve and be sure to state any nonzero values ofthat quantity at y 0 andy mm. In other words, label the vertical axis when you can.) -1mm V+1000 v mim mim Electric Field (x-component) Electric Field (y-component) Electric Potential (V 0aty 0) Electric Potential (function of y) For the third graph above, pick a y (state your y value) and plot V as a function of x as you move 1 mm horizontally in the capacitor For the graphs below, draw one plot for a proton (graphs 1 and 3) and one for an electron (graphs 2 and 4). Electric Force (x-component) Electric Force (y-component) Electric PE (function of x) Electric PE (function of y)

Answer 1

E^rightarrow = -nabla^rightarrow V = -[x^partial differential V/partial differential x + y^partial differential v/y + z^partial differential v/partial differential z = + 1000 v/max therefore E_x = 0 & E_y = -1000 y^= 1000 v/max therefore because partial differential v/partial differential x = 0 implies v(x) = constant & partial differential v/partial differential y = 1000 v(y) = (1000)y implies v(y) alpha y F_x = qE_x = 0 (E_x = 0) -partial differential v/partial differential x = 0 implies V(x) = constant F_y = qE_y = eE_y = -e(1000) y^(for proton) = -eE_y = -e(-1000)y^= 1000 y^(for e^-)} -partial differential v/partial differential y = (1000e) implies v(y) = (1000)y for proton -partial differential v/partial differential y = (1000e) implies v(y) = -(1000 e)y for e^-