A copper penny has a mass of 3.0 g. A total of 4.0 × 10^12 electrons are transferred from one neutral penny to another. If the e

Question

2 years ago

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lectrostatic force of attraction between the pennies is equal to the weight of a penny, what is the separation between them?

2 answers:

olga55 [171] · Answer 1 · 2023-08-26T00:10:13+03:00

Then the separation between the two pennies will be O.354m.

To find the correct answer, we need to know about the electrostatic force of attraction.

<h3>What is the separation between the pennies?</h3>

$m=3*10^{-3}kg\\n=4*10^{12}\\$

As we know that the charge of one electron is 1.6×10^-19C.Thus, the total charge on the penny will be,

$Q_1=4*10^{12}*1.6*10^{-19}=6.4*10^{-7}C.\\Q_2=-Q_1$

$F_e=\frac{kQ_1Q_2}{r^2} =mg\\$ , here given that the electrostatic force equal to the weight of the system.

$r=\sqrt{\frac{kQ_1Q_2}{mg} } =\sqrt{\frac{(6.4*10^{-7})^2}{4*3.14*8.85*10^{-12}*3*10^{-3}*9.8} }\\\\r=0.354m$

Thus, we can conclude that the separation between the pennies is 0.354m.

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lbvjy [14] · Answer 2 · 2023-08-26T02:12:23+03:00

0.354m will be the separation between the two pennies.

We must understand the electrostatic force of attraction in order to arrive at the correct solution.

<h3>How far apart are the pennies from one another?</h3>

One electron has a charge of 1.6 10-19 coulombs, as is common knowledge. Consequently, the total fee for the penny will be,

$Q_1=ne=4*10^{12}*1.67*10^{-19}=6.4*10^{-7}C\\Q_2=-Q_1$

$F=\frac{kQ_1Q_2}{R^2}=mg$

Given that the electrostatic force in this instance is equal to the system's weight.

$R=\sqrt{\frac{kQ_1Q_2}{mg} } =0.354m$

As a result, we may say that the distance between the pennies is 0.354m.

Learn more about electrostatic force here:

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