Question #4b856

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Question #4b856

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Answer 1 · 2017-04-25T14:51:52

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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:

$\int \int_{S} E \cdot d S = \frac{\sum Q_{e n c}}{ε_{o}}$ $int int_S mathbf E cdot d mathbf S = (sum Q_(enc))/epsilon_o$

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

$E = \frac{\sum Q_{e n c}}{4 π ε_{o} r^{2}}$ $mathbf E = (sum Q_(enc))/( 4 pi epsilon_o r^2)$

For these conducting spheres, we have 2 different situations:

inside the sphere, $\sum Q_{e n c} = 0$ $sum Q_(enc) = 0$ so $E = 0$ $mathbf E = mathbf 0$
oustide the sphere, $\sum Q_{e n c} = Q$ $sum Q_(enc) = Q$ so $E = \frac{Q}{4 π ε_{o} r^{2}} e_{r}$ $mathbf E = ( Q)/( 4 pi epsilon_o r^2) mathbf e_r$

We connect electric field $E$ $mathbf E$ to potential via $E = - \nabla V$ $mathbf E = - nabla V$ , which simplifies due to symmetry to $E = - \frac{\partial V}{\partial r} e_{r}$ $mathbf E= - (partial V)/(partial r) mathbf e_r$ .

So we get these results:

inside the sphere, $E = 0 \Rightarrow V = const$ $mathbf E = mathbf 0 implies V = " const"$
outside the sphere, $\sum Q_{e n c} = Q$ $sum Q_(enc) = Q$ so $E = \frac{Q}{4 π ε_{o} r^{2}} e_{r} \Rightarrow V = \frac{Q}{4 π ε_{o} r}$ $mathbf E = ( Q)/( 4 pi epsilon_o r^2) mathbf e_r implies V = ( Q)/( 4 pi epsilon_o r)$

Because we know $V$ $V$ 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:

$V_{1} = = \frac{Q_{1}}{4 π ε_{o} R_{1}}$ $V_1 = = ( Q_1)/( 4 pi epsilon_o R_1)$

Conservation of charge tells us that $Q_{2} = 2 Q_{1}$ $Q_2 = 2 Q_1$ . Conservation of matter (volume) tells us that: $R_{2} = \sqrt[3]{2} R_{1} \approx 1.26 R_{1}$ $R_2 = root3(2) R_1 approx 1.26 R_1$ .

So we can say that:

$V_{2} = \frac{Q_{2}}{4 π ε_{o} R_{2}}$ $V_2 = ( Q_2)/( 4 pi epsilon_o R_2)$

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

$= root3(4) V_1$

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

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Question #4b856

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