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3 months ago

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A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int

o n identical spherical droplets, each of which has electric potential Vsmall throughout its volume. The n small droplets are far enough apart from one another that they do not interact significantly.
Find (Vbig)/(Vsmall) , the ratio of Vbig, the electric potential of the initial drop, to Vsmallthe electric potential of one of the smaller drops.(The ratio should be dimensionless and should depend only on n)

1 answer:

kicyunya [3.2K]3 months ago

7 0

Answer:

$\frac{V_{big}}{V_{small}} = n^{2/3}$

Explanation:

Let the charge on the large droplet be denoted as Q.

When the radius of the droplet is R, the electric potential for the larger droplet can be expressed as:

$V_{big} = \frac{KQ}{R}$

If it splits into n identical droplets, let the charge of each be "q" and their radius be "r".

Applying volume conservation gives us:

$\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3)$

$r = \frac{R}{n^{1/3}}$

Now, the potential for the smaller droplets is given as:

$V_{small} = \frac{kq}{r}$

$V_{small} = \frac{K(Q/n)}{\frac{R}{n^{1/3}}}$

$V_{small} = \frac{1}{n^{2/3}}\frac{KQ}{R}$

$\frac{V_{big}}{V_{small}} = n^{2/3}$

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Explanation:

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Complete Question

An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.

At time 5 seconds, I = 1.2 A.

We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

The question indicates that

The wire’s diameter is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

Aluminum's resistivity is $2.75*10^{-8} \ ohm-meters.$

The electric field variation is described as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

The charge is effectively given by the equation

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area expressed as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

Thus,

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

By substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

The question states that t = 5 seconds

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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