Homework Help

10^6 spherical conducting droplets are combined to create a single drop. The radius of...

user profile pic

leevicboy | (Level 1) eNoter

Posted September 30, 2013 at 4:08 AM via web

dislike 1 like

10^6 spherical conducting droplets are combined to create a single drop. The radius of each droplet is r=5 micrometer with a charge of q=1.6*10^-14C. How much energy is needed to overcome the Coulomb's repulsive force in creating this drop?

1 Answer | Add Yours

user profile pic

valentin68 | College Teacher | (Level 3) Associate Educator

Posted September 30, 2013 at 5:33 AM (Answer #1)

dislike 2 like

If we have a system formed by `n` particles, the potential energy of the entire system is the sum over `C_n^2 =(n(n-1))/2` (combinations of n elements taken as 2) pairs of the potential energy of a single pair of two particles.

The electric potential of a single droplet at distance `R` is

`U = k_e*Q/R =9*10^9*(1.6*10^-14)/(5*10^-6) =28.8 V`

To form a two particle system we need to bring the second particle from infinity to distance R. Thus the potential energy of the two particle system is

`E_p = q*U =1.6*10^-14*28.8 =4.608*10^-13 J`

Now for `n=10^6` particles the total potential energy is

`E_("tot") =(n*(n-1))/2 *E_p =(10^6)^2/2 *4.608*10^-13 =0.2304 J`

Thus the total energy needed to overcome the Coulomb repulsive force, in the creation of the final drop is 0.2304 J

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes