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?
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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
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