The laboratory test that detects neutropenia is

n isolated charged soap bubble of radius R0=7.45 cmR0=7.45 cm is at a potential of V0=307.0 volts.V0=307.0 volts. If the bubble

Gnesinka [82]

Complete Question

An isolated charged soap bubble of radius R0 = 7.45 cm is at a potential of V0=307.0 volts. V0=307.0 volts. If the bubble shrinks to a radius that is 19.0%19.0% of the initial radius, by how much does its electrostatic potential energy ????U change? Assume that the charge on the bubble is spread evenly over the surface, and that the total charge on the bubble r

Answer:

The difference is $U_f -U_i = 16 *10^{-7} J$

Explanation:

From the question we are told that

The radius of the soap bubble is $R_o = 7.45 \ cm = \frac{7.45}{100} = 0.0745 \ m$

The potential of the soap bubble is $V_1 =307.0 V$

The new radius of the soap bubble is $R_1 = 0.19 * 7.45=1.4155\ cm = 0.014155 \ m$

The initial electric potential is mathematically represented as

$U_i = \frac{V_1^2 R_o }{2k }$

The final electric potential is mathematically represented as

$U_f = \frac{V_2^2 R_1 }{2k }$

The initial potential is mathematically represented as

$V_1 = \frac{kQ}{R_o}$

The final potential is mathematically represented as

$V_2 = \frac{kQ}{R_1}$

Now

$\frac{V_2}{V_1} = \frac{R_o}{R_1}$

substituting values

$\frac{V_2}{V_1} = \frac{7.45}{1.4155} = \frac{1}{0.19}$

=> $V_2 = \frac{V_1}{0.19}$

So

$U_f = \frac{V_1^2 R_2 }{2k * 0.19^2}$

Therefore

$U_f -U_i = \frac{V_1^2 R_2 }{2k * 0.19^2} - \frac{V_1^2 R_o }{2k }$

$U_f -U_i = \frac{V_1^2}{2k} [\frac{ R_1 }{ * 0.19^2} - R_o]$

where k is the coulomb's constant with value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$U_f -U_i = \frac{307^2}{9 * 10^{9}} [\frac{ 0.014155 }{ 0.19^2} - 0.0745]$

$U_f -U_i = 16 *10^{-7} J$

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Attached is the solution

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<h3><u>Freefall</u></h3>

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