Respuesta :
Answer:
Charge, [tex]q' = -3.29\times 10^{-6} C[/tex]
Explanation:
The added charge must create the force which is both equal and opposite to the charge that is caused by 3μC in the lower right corner.
Now we know that the electrostatic force of a charge is given by
[tex]F=k.\frac{q^{2}}{(4d)^{2}}[/tex]
[tex]F=k.\frac{q^{2}}{16d^{2}}[/tex]
We know
Electrostatic force between the charges at the corner is
[tex]F'=-k.\frac{qq'}{17d^{2}}.\frac{4d}{d\sqrt{17}}[/tex]
Then to find the charge, q' we get F=F'
[tex]k.\frac{q^{2}}{16d^{2}}[/tex] = [tex]-k.\frac{qq'}{17d^{2}}.\frac{4d}{d\sqrt{17}}[/tex]
[tex]\frac{q}{16} = -\frac{q'}{17}.\frac{4}{\sqrt{17}}[/tex]
[tex]q' = -\frac{17q}{16}\sqrt{\frac{17}{4}}[/tex]
Now substituting q = 3μC. we get
[tex]q' = -\frac{17\times 3 \mu C}{16}\sqrt{\frac{17}{4}}[/tex]
[tex]q' = -3.29\times 10^{-6} C[/tex]
The charge that must be placed at the empty corner is : q' = -3.29 * 10⁻⁶ C
Given that the Electrostatic force of a charge is expressed as
F = [tex]K . \frac{q^{2} }{(4d)^2}[/tex] ---- ( 1 )
The added charge ( q' ) will create a charge equal and opposite to the charge ( q ) caused by 3μC
∴ The electrostatic force between the charges at the corner is :
F' = [tex]-k . \frac{qq'}{17d^2} * \frac{4d}{d\sqrt{17} }[/tex] ----- ( 2 )
Final step : The charge that must be placed at the empty corner ( q' )
where F = F'
Equating equations ( 1 ) and ( 2 )
[tex]q' = - \frac{17q}{16} \sqrt{\frac{17}{4} }[/tex] ------ ( 3 )
Insert q = 3μC in equation ( 3 )
q' = -3.29 * 10⁻⁶ C
Hence we can conclude that the charge that must be placed at the empty corner is : q' = -3.29 * 10⁻⁶ C
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