Physics, 05.06.2020 02:57, hectorav6619
A 12-V battery is connected to an air-filled capacitor that consists of two parallel plates,
each plate has an area of 7.60 cm2
. The separation between the plates of the capacitor is d =
0.30 cm. (Assume the electric field between the plates to be uniform).
a. Draw the situation.
b. Find the magnitude of the electric field between the plates.
Now, a proton (q = 1.6 x10-19 C) is released from rest at the positive plate of the capacitor.
c. Calculate the electric potential energy gained by the proton just before it touches the negative
plate.
A slab of Teflon of dielectric constant k =2.1 is then inserted between the plates of the
capacitor.
d. What is the new capacitance of the capacitor?
e. Calculate the change in the total energy stored in the capacitor before and after inserting the
dielectric slab.
Answers: 2
Physics, 23.06.2019 15:00, Zaayyyy
An experiment was performed on a 2-kg block. forces of 5, 10 and 15 newtons respectively were applied to the block for 5 seconds. describe the difference in acceleration between the three trials. a) the final accelerations will all be the same. b) the final acceleration will triple between the first and third trial. c) the final acceleration will be cut by 1/3 from the first to third trial. d) the final acceleration will increase nine time between the first and third trial.
Answers: 1
A 12-V battery is connected to an air-filled capacitor that consists of two parallel plates,
each...
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