Please draw pictures!! 56. (I) A parallel-plate capacitor has fixed charges+Q and -Q. The separation of...
Constants Part A A parallel-plate capacitor has fixed charges +Q and-Q By what factor does the energy stored in the electric field change? Express your answer using two significant figures. PEfinal PEjnitia Previous Answers Request Answer Submit X Incorrect, Try Again, 4 attempts remaining ▼ Part B How much work must be done to reduce the plate separation from d to 2 d? The area o
3. A parallel-plate capacitor with plate separation x and area A has charges +Q and -Q on its plates. The capacitor is disconnected from the source of charge, so the charge on plates stays constant. a) What is the total energy stored in the capacitor? b) The plates are pulled apart by distance dx. What is the change in stored energy?
If the area of the plates of a parallel plate capacitor is halved and the separation between the plates tripled, while the charge on the capacitor remains constant, then by what factor does the energy stored in the capacitor change? decreases by a factor of 2/3 increases by a factor of 6 decreases by a factor of 1/6 increases by a factor of 3/2 increases by a factor of 2
A parallel-plate capacitor (Homework #2) A parallel-plate capacitor is made by a pair of plates with area A and separation d. At first, the plates are charged up to +Q and -O respectively, and then a battery is disconnected. Express all the answers for a) through f) only by Q, d, A and 5 +Q 2d -Q Area A a) What is the electric field E inside of the capacitor? b) What is the potential difference V between the plates?...
5 A parallel-plate capacitor (Homework #2) A parallel-plate capacitor is made by a pair of plates with area A and separation d. At first, the plates are charged up to+ and - respectively, and then a battery is disconnected. Express all the answers for a) through f only by Q, d, A and s 2d Area A Now the places are pulled apart until the separation becomes 2d What is the electric field E'inside of the capacitor? (Express E" by...
plz take into account sig figures and make sure its right i have few chances left :) A parallel-plate capacitor has fixed charges + and-Q. The separation of the plates is then halved. By what factor does the energy stored in the electric field change? Express your answer using two significant figures. PEfinal PEinitial = 0.50 Submit Previous Answers Correct Part B How much work must be done to reduce the plate separation from d to zd? The area of...
A parallel plate capacitor has a charge Q, plates of area A and separation d, where , so that we can disregard the fringe field. Show that the magnitude of the force exerted on each plate by the other is F = Q2/(2e0A).You may use (with care!) the expression F = QE, or consider the change in potential energy of the capacitor as you change the separation of the plates. Or you may do both.
what is b plz . Consider a disconnected parallel plate capacitor with area Ao and width --4, 3do. There is a magnitude of charge density g, on each plate. For this problem you may assume the plates are large enough to neglect fringing fields. 3d How much energy is stored in the capacitor?2 z a. For the purpose of simplifying the math, assign this value 3E. A large conducting slab of area A and width do is now placed within...
A parallel plate capacitor is constructed with circular plates of radius 0.750 cm and plate separation 0.0500 mm. If the capacitor is connected across a 37.2 V source, find: a) the capacitance b) the surface charge on each plate c) The energy stored in the capacitor d) the electric field between the plates e) the energy density between the plates
A parallel-plate capacitor is charged by being connected to a battery and is kept connected to the battery. The separation between the plates is then doubled. How does the electric field, charge and total energy change? Electric field is halved, charge is doubled, total energy is doubled Electric field is doubled, charge is halved, total energy is doubled Electric field is doubled, charge is halved, total energy is halved Electric field is halved, charge is halved, total energy is halved