Consider the diagram below. Which plate (left or right is the positive plate? Ho
ID: 1536822 • Letter: C
Question
Consider the diagram below. Which plate (left or right is the positive plate? How do you know? What is the electric field strength inside the capacitor? What is the change in electric potential energy if a proton is moved from the right plate to a position at a location midway between the plates. Does it matter how you get it to that position? Consider the two charges equal in magnitude and opposite in sign shown below. Draw the electric field in the region around (mostly between these two opposite charges. Draw at least 3 equipotential surfaces near these charges. Explain how you located the equipotential surfaces.Explanation / Answer
4, a) The poitive plate of a capacitor is always at a higher plate thanthe negative plate hence the right plate (-300 V) is +ve ly charged
b) Electric field, E = dV/d
E = (-300 - (-600))/(3*10^-3) = 300,000/3 = 10^5 V/m
c) Potential energy of a charged particle at any point in spzace = qV ( where q ia the chaege of the particle andthe V is the electric potential at that point in space)
hence change in PE from position st the right plate to the position at midway between the plates is
PE t right plate - PE midway = qVr - qVm = q(-300 - (-450)) [ here Vm = -450 V because it can be seen that the potential is linearlky falling from -300 to 0 600 in regular steps, hence at the middle of the capacitor is is exactly in the middle of -300 to -600 V, i.e. 400 V]
dPE = 150q Joules
dPE = 150*1.6*10^-19 = 2.4*10^-18 J
It does not mwatter how we arrive at that position because the electrostatic force is a conservative forece and conservative forces hae the property that potential enegy in a conservative is only deoeondent on the inidial position and the final position and notht heroute taken to reach there