Question

Consider three identical spherical conductors. One conductor, labeled A, is given a charge of 7µC of type Nell. The charge state of the second and third conductors, B and C, are unknown. The conductors undergo a sequence of 3 contacts that occur ONLY between pairs of conductors. After the contacts, the net charge on each conductor is A: (0.25µC of type Nell), B: (1.875µC of type Nell) and C: (1.875µC of type Nell).

What was the initial charge state of each sphere? What is the sequence of contacts? How many electrons were transferred from each sphere to the other in each contact event? Make the assumption that the type Paul charge state corresponds to an excess of electrons.

Answer 1

When sphere are coming contact then both sphere will be at same potential, and since sphere are identical then each sphere will have equal charge after contact to make same potential.

Let B and C have x and y charges respectively

So, when A and B come in contact then charge is equally divided to both sphere,

so, A and B both have charge of =(x+7)/2.

Now then A and C are coming in contact the charge on A and C both have charge =((x+7)/2 +y)/2

so, after contact from B and C,

Final charge on A=((x+7)/2 +y)/2 =0.25

=> x+7+2y=0.5------------------------------------------------eq(1)

now when B and C come in contact then charge on each sphere will be= (((x+7)/2 +y)/2 +(x+7)/2)/2

So final charge on B after contact with A and C =(((x+7)/2 +y)/2 +(x+7)/2)/2 =(2x +2y +14)/4

1.875 =(2x +2y +14)/4

x+y+7=3.75 ----------------------------------------------------eq(2)

to get the value of x and y we need to solve eq(1) and eq(2) together(subtract eq(2) from 1)

y =0.5-3.75=-3.25 uC

and x =3.75+3.25-7 =0 uC ( no charge on sphere B)

part(2)

sequence of contact=> A &B , A&C, and B and C

part(3)

once you get the charge you can calculate the no of electron transfer using formula =charge tranfer/(charge on a electron).

charge on 1 electron = 1.6 *10^(-19) C