Potential on merging water droplets

[Alan I]

Homework Statement

Suppose you have two identical droplets of water, each carrying charge 2.8 pC spread uniformly through their volume. The potential on the surface of each is 589 Volts.

Now, you merge the two drops, forming one spherical droplet of water. If no charge is lost, find the potential at the surface of this new large water droplet. in V.

Homework Equations

V = kQ/r
ρ=Q/V

The Attempt at a Solution

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I wasn't sure how to approach this problem so I assumed that the total volume will be conserved (since water is not compressible) and therefore if the volume of one initial droplet doubles after merging (V=2XV₀) then the radius will increase by 1/2Xr₀.

So by following that reasoning I got r_final = 6.41 x 10^-5 m

⇒ V_final = k*Q_total / r_final

⇒ Taking the total conserved charge to be Q₁ + Q₂ ⇒ (2)*(2.8 X 10^-12) = 5.6 x 10^-12 C

Plugging in all the values I got V_final = 785 V , and this result turned out to be wrong

 

[gleem]

I would look at the value of the radius of the coalesced drop that you calculated.

The volume of a sphere is 4/3 π R³

 

[Alan I]

I would look at the value of the radius of the coalesced drop that you calculated.

The volume of a sphere is 4/3 π R³

Thanks, now I see it.

So from V_droplet = k (2.8x10^-12)/r₀,
I got r₀ = 4.27 x 10^-5

Plugged that in V₀ = 4/3*π*(r₀)³ and got V₀ = 3.27 x 10^-13

Now Vol_final = 2V₀ = 6.54 x 10^-13

⇒ 6.54 x 10^-13 = 4/3*π*R³

⇒ R = (1.56 x 10^-13)^1/3

⇒ V_{big droplet} = k Q_total/R

Thanks!