Question

Question #4b856

Answer 1

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 mathbf E cdot d mathbf S = (sum Q_(enc))/epsilon_o`

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

`mathbf E = (sum Q_(enc))/( 4 pi epsilon_o r^2)`

For these conducting spheres, we have 2 different situations:

• inside the sphere, `sum Q_(enc) = 0` so `mathbf E = mathbf 0`

• oustide the sphere, `sum Q_(enc) = Q` so `mathbf E = ( Q)/( 4 pi epsilon_o r^2) mathbf e_r`

We connect electric field `mathbf E` to potential via `mathbf E = - nabla V`, which simplifies due to symmetry to `mathbf E= - (partial V)/(partial r) mathbf e_r`.

So we get these results:

• inside the sphere, `mathbf E = mathbf 0 implies V = " const"`

• outside the sphere, `sum Q_(enc) = Q` so `mathbf E = ( Q)/( 4 pi epsilon_o r^2) mathbf e_r implies V = ( 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 = = ( 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 = root3(2) R_1 approx 1.26 R_1`.

So we can say that:

`V_2 = ( Q_2)/( 4 pi epsilon_o R_2)`

`= ( 2 Q_1)/( root3(2) \ 4 pi epsilon_o R_1)`

`= root3(4) V_1`

`= root3(4) cdot 250 " V" approx 397 " V"`