# electric fields help

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#1
Its question 1c on here

https://pmt.physicsandmathstutor.com...s%201%20QP.pdf

https://pmt.physicsandmathstutor.com...s%201%20MS.pdf

these three bullet points from the answer i dont understand

It does not reach plate B / It reverses direction
When it returns to A it has 4 eV (of KE)
It stops 2/3 of the distance across the plates (AW)

How were these worked out?
0
2 years ago
#2
Direction of electric field is from A to B
Force on a positive charge is along the direction of the field, i.e along line AB
Charge on an electron is negative
So the force on electron is from B to A, opposite to the direction of motion.
There is a ******ing force. The electron will decelerate.

Using loss in KE = gain in electric PE
The minimum KE needed for an electron to reach B from A is
loss in KE = qV = 6 eV
But the electron only has 4 eV, so it does not reach plate B.
It only has 2/3 of the energy required, so it reaches 2/3 along line AB. (v^2 = 2as, so s is proportional to kinetic energy)

When it returns to A, it covers 2/3 of the distance AB, so the work done by the resultant electric force on the electron is 2/3 x 6 = 4 eV, hence it has 4 eV of KE when it returns to A.
0
#3
thank you so much
(Original post by BobbJo)
Direction of electric field is from A to B
Force on a positive charge is along the direction of the field, i.e along line AB
Charge on an electron is negative
So the force on electron is from B to A, opposite to the direction of motion.
There is a ******ing force. The electron will decelerate.

Using loss in KE = gain in electric PE
The minimum KE needed for an electron to reach B from A is
loss in KE = qV = 6 eV
But the electron only has 4 eV, so it does not reach plate B.
It only has 2/3 of the energy required, so it reaches 2/3 along line AB. (v^2 = 2as, so s is proportional to kinetic energy)

When it returns to A, it covers 2/3 of the distance AB, so the work done by the resultant electric force on the electron is 2/3 x 6 = 4 eV, hence it has 4 eV of KE when it returns to A.
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