Question

Problem 2 A charge O is tran ferred rom an initially uncharged plastic ball to an entical ball 13.50 ㎝ away. The force of attraction is then 19.50 my How many electrons were transferred rom one ball to the other? Submit Answer Tries 0/6

Answer 1

Force of attraction,

$\large F=\frac{kq_1q_2}{r^2}$

Here, q1=q2=Q

$\large \therefore F=\frac{kQ^2}{r^2}$

Solving the equation for Q, we get,

$\large Q=\sqrt\frac{Fr^2}{k}$

● Givens :

$\large r=13.5\,cm=13.5\times10^-^2\,m$

  $\large F=19.5\,mN=19.5\times10^-^3\,N$

Now,

$\large Q=\sqrt\frac{19.5\times10^-^3\times(13.5\times10^-^2)^2}{9\times10^9}=1.987\times10^-^7\,C$

We know that,

Charge = number of electrons × electronic charge

Q = ne

Therefore,

Number of electrons,

  $\large n=\frac{Q}{e}=\frac{1.987\times10^-^7\,C}{1.6\times10^-^19}=1.24\times10^1^2$