# Question #4b856

Apr 25, 2017

I think we have to model these as charged conducting spheres so the charges are at the surface as shown .

Gauss' Law tells us that:

$\int {\int}_{S} m a t h b f E \cdot d m a t h b f S = \frac{\sum {Q}_{e n c}}{\epsilon} _ o$

And by using a concentric Gaussian sphere, we can say of a general sphere of radius $r$ that:

$m a t h b f E = \frac{\sum {Q}_{e n c}}{4 \pi {\epsilon}_{o} {r}^{2}}$

For these conducting spheres, we have 2 different situations:

• inside the sphere, $\sum {Q}_{e n c} = 0$ so $m a t h b f E = m a t h b f 0$

• oustide the sphere, $\sum {Q}_{e n c} = Q$ so $m a t h b f E = \frac{Q}{4 \pi {\epsilon}_{o} {r}^{2}} m a t h b f {e}_{r}$

We connect electric field $m a t h b f E$ to potential via $m a t h b f E = - \nabla V$, which simplifies due to symmetry to $m a t h b f E = - \frac{\partial V}{\partial r} m a t h b f {e}_{r}$.

So we get these results:

• inside the sphere, $m a t h b f E = m a t h b f 0 \implies V = \text{ const}$

• outside the sphere, $\sum {Q}_{e n c} = Q$ so $m a t h b f E = \frac{Q}{4 \pi {\epsilon}_{o} {r}^{2}} m a t h b f {e}_{r} \implies V = \frac{Q}{4 \pi {\epsilon}_{o} r}$

Because we know $V$ to be constant within the sphere, we can say that, at the surface and inside each of the initial waterdrops we have potential as follows:

${V}_{1} = = \frac{{Q}_{1}}{4 \pi {\epsilon}_{o} {R}_{1}}$

Conservation of charge tells us that ${Q}_{2} = 2 {Q}_{1}$. Conservation of matter (volume) tells us that: ${R}_{2} = \sqrt[3]{2} {R}_{1} \approx 1.26 {R}_{1}$.

So we can say that:

${V}_{2} = \frac{{Q}_{2}}{4 \pi {\epsilon}_{o} {R}_{2}}$

$= \frac{2 {Q}_{1}}{\sqrt[3]{2} \setminus 4 \pi {\epsilon}_{o} {R}_{1}}$

$= \sqrt[3]{4} {V}_{1}$

$= \sqrt[3]{4} \cdot 250 \text{ V" approx 397 " V}$