Oppositely charged parallel plates are separated by 3.92 mm. A potential difference of 600 V exists between the plates.
(a) What is the magnitude of the electric field between the plates?
(b) What is the magnitude of the force on an electron between the plates?
(c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.69 mm from the positive plate?
(a) A uniform electric field of magnitude E, directed perpendicular to the plates, exists in the region between the plates and exerts a constant force F = qE on a charged particle in this region.
When the particle moves a distance d parallel to the field, going from one plate to the other, the work done by the field is W = Fd cos 0 = qEd.
Thus, the magnitude of the change in potential energy of the particle is |?PE| = W = qEd,
and the magnitude of the potential difference between the plates is |?V| = |?PE|/q
We know the potential difference between the plates and the distance d = 3.92 mm between the plates. Therefore, the magnitude of the electric field between the plates is
(b) The magnitude of the force on an electron of charge magnitude e located between the plates is
(c) Given that the distance between the plates is d = 3.92 mm, the displacement s of the electron as it moves from its initial position of 2.69 mm to the negative plate is
An applied force with magnitude F = eE, directed toward the negative plate, must be used to offset the influence of the field, moving the electron without acceleration. The work done by the applied force as the electron is moved to the negative plate is
Please show how your came about solving each THank you!!!
= xxx/x.xx*xx^-x = 1.53061*10^5 x/x
x) x = E*q = 1.53061*10^5 * x.x*xx^-xx = 2.4489*10^-14 x
x) xxxx = F*x = 2.4489*10^-14 x * (x.xx-x.xx)*.xxx= 3.0122 * 10^-17 x xxxx 1
(x) A xxxxxxx xxxxxxxx xxxxx of xxxxxxxxx E, xxxxxxxx xxxxxxxxxxxxx xx the xxxxxx, exists xx xxx xxxxxx between xxx plates xxx xxxxxx x constant xxxxx F = xx xx a xxxxxxx particle xx xxxx xxxxxx. When xxx particle xxxxx x xxxxxxxx d xxxxxxxx to xxx xxxxx, xxxxx from xxx plate xx xxx xxxxx, the xxxx done xy xxx xxxxx is x = xx xxx x° = xxx. Thus, xxx xxxxxxxxx xx the xxxxxx in xxxxxxxxx xxxxxy xx the xxxxxxxx is |?xx| = x = xxx, and xxx xxxxxxxxx xx the xxxxxxxxx difference xxxxxxx xxx xxxxxx is
xx xxxx xxx potential xxxxxxxxxx between xxx xxxxxx xxx the xxxxxxxx d = x.xx xx between xxx plates. xxxxxxxxx, xxx xxxxxxxxx of xxx electric xxxxx xxxxxxx xxx plates xx E = x.xxxxx*xx^x x/x
x = x.xxxx*xx^-xx N xxxx 3
(x) xxx xxxxxxxxx of xxx force xx xx xxxxxxxx of xxxxxx magnitude x xxxxxxx xxxxxxx the xxxxxx is
(x) xxxxx xxxx the xxxxxxxx between xxx xxxxxx xx d = 3.92 xx, xxx xxxxxxxxxxxx s xx the xxxxxxxx xx xx moves xxxx its xxxxxxx xxxxxxxx xx 2.69 xx to xxx xxxxxxxx xxxxx is x = x.xx-x.xx xx = 1.23 xx
xx xxxxxxx xxxxx with xxxxxxxxx F = xx, xxxxxxxx toward xxx negative xxxxx, xxxx xx used xx offset xxx xxxxxxxxx xx the xxxxx, moving xxx xxxxxxxx xxxxxxx acceleration. xxx work xxxx xy xxx applied xxxxx as xxx xxxxxxxx xx moved xx the xxxxxxxx xxxxx xx W =x.xxxx * xx^-xx x