Physics: Two particles are fixed on an x axis. particle 1 of charge 69.9 μc is located at x = -4.32 cm; particle 2 of charge q is located at x = 36.1 cm. particle 3 of charge magnitude 22.1 μc is released from rest on the y axis at y = 4.32 cm. what is the value of q if the initial acceleration of particle 3 is in the positive direction of (a) the x axis and (b) the y axis?

Two particles are fixed on an x axis. particle

Ответ:

Tok

02.01.2022

Physics

a) $1.45\times 10^{-2}\ \rm C$

b) $1.76\times10^{-3}\ \rm C$

Explanation:

Given:

Charge on particle(1) = $69.9\ \mu C$

Charge on particle(3) = $22.1\ \mu C$

Angle made by particle 1 with the x axis

$\tan \theta=\dfrac{4.32}{4.32}\\\theta =45^\circ$

Angle made by particle 3 with the x axis

$\tan \theta=\dfrac{4.32}{36.1}\\\theta =6.82^\circ$

a) If the acceleration particle 3 is along the positive x axis then then it means the net force on the charge along y axis is zero so

$\dfrac{k\times 69.9}{r_1^2}\times \sin45^\circ=\dfrac{k\times Q}{r_2^2}\times \sin6.28^\circ\\\dfrac{k\times 69.9}{37.32}\times \sin45^\circ=\dfrac{k\times Q}{1321.8}\times \sin6.28^\circ\\Q=1.45\times10^{-2}\ \rm C$

b) If the acceleration particle 3 is along the positive y axis then then it means the net force on the charge along x axis is zero so

$\dfrac{k\times 69.9}{r_1^2}\times \cos45^\circ=\dfrac{k\times Q}{r_2^2}\times \cos6.28^\circ\\\dfrac{k\times 69.9}{37.32}\times \cos45^\circ=\dfrac{k\times Q}{1321.8}\times \cos6.28^\circ\\\\Q=1.76\times10^{-3}\ \rm C$

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Ответ:

lclaudettecarte4720

06.11.2021

Physics

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Limitations:

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Explanation:

For more information:

https://www.who.int/water_sanitation_health/dwq/gdwqvol32ed.pdf?ua=1https://www3.epa.gov/region1/eco/drinkwater/your_drinkwater_source.html

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