Basic Potential Energy Question. Kinda Tricky! (Check my Work)?

Question

Question: a plate carries a charge of -3.0 uC, while a rod carries a charge of +2.0 uC. How many electrons must be transferred from the plate to the rod, so that both objects have the same charge?

My Approach:

N= q/e = -2.0e-6/-1.6e-19 = 1.25e13 electrons

What is the other part asking? does it include the -3.0 uC? That's where im stuck...my prof marked me wrong for having -2.0e-6 (without the units) - so am i technically right if i have the units attached to -2.0e-6 C? Thanks...

Answer 1

Some answerers are presuming that they are touched, and then question is framed. They are not capacitors. In case they are touched then flow will continue to take place till both become equi-potential surface.

It is asked how many electrons should be transferred.

What is neutral surface: Net Charge = 0

If you knock out 1 electron (e=1.6*10^-19 C) from a surface then it acquires a net positive charge of (+ e) i.e. absence of an electron.

It means a net positive charge = absence or shortfall of electrons

Rod = Net Positive charge= + 2*10^-6 C

Rod is short in electrons = 2*10^-6 C / 1.6*10^-19 C =
Nr = 1.25*10^13 =12500*10^9

Plate = Net Negative charge = 3*10^-6 C

Plate is surplus in electrons = 3*10^-6 C / 1.6*10^-19 C
Np= 1.875*10^13 =18750*10^9

So plate will transfer to rod.

Given: upon transfer, charge on each becomes same say Q=(q1+q2) / 2
OR Number of free electrons on each become same N = (Nr+Np)/2

Each should have N=15625*10^9 free electrons.

Plate will transfer = Np – N = (18750-15625)*10^9 = 3125*10^9 electrons

After transfer

The Rod will have = Nr + N = (12500+3125)*10^9 = 15625*10^9 electrons
With plate = 15625*10^9 electrons

Answer 2

The total charge on both elements is -1.0µC. For the system to end up with both having the same charge, this amount must be shared equally between the elements. Therefore the resulting charge on each element will be -.5µC. -2.5µC will be transferred from the -3µC plate to the 2µC rod so that it ends up with -.5µC. The number of electrons transferred is then

-2.5*10^-6/-1.6*10^-19 = 1.563*10^13

Answer 3

Agree with gp4rts.