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Sagot :
A third positive charge [tex]x = \frac{2\sqrt{2}a }{1+\sqrt{2} }[/tex] can be placed on the x-axis so that the net electric force on it is zero.
whilst -factor charges are in touch, there may be
[tex]F = k \frac{q_{1}q_{2} }{ r^{2} }[/tex]
Where
[tex]k = 9 * 10^{9} Nm^{2}C^{-1}[/tex] = the coulomb constant.
The charges' magnitudes are [tex]q_{1}[/tex] and [tex]q_{2}[/tex].
r is the separation of the charges.
Let,
The third charge equals Q at, x equals x from the source.
because there is no net force acting on the third charge.
Hence,
[tex]f_{1} + f_{2} = 0[/tex]
⇒ [tex]k\frac{(2q)Q}{x^{2} }-k\frac{qQ}{(2a-x)^{2} } = 0[/tex]
⇒ [tex]\frac{2}{x^{2} } - \frac{1}{(2a-x)^{2} } = 0[/tex]
⇒ [tex]\frac{2}{x^{2} } = \frac{1}{(2a-x)^{2} }[/tex]
⇒ [tex]\frac{\sqrt{2} }{x} = \frac{1}{2a-x}[/tex]
⇒ [tex]2\sqrt{2}a - \sqrt{2} = x[/tex]
⇒ [tex]x = \frac{2\sqrt{2}a }{1+\sqrt{2} }[/tex]
handiest when forces resulting from previous prices cancel every different out will the pressure on a third fee be 0.
at the same time as the pressure between identical costs is repulsive and is interpreted as high-quality, the pressure between two opposing expenses is attracted and is interpreted as bad.
Know more about particles of charge:
https://brainly.com/question/18750504
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