Get expert insights and reliable answers to your questions on IDNLearn.com. Discover detailed answers to your questions with our extensive database of expert knowledge.
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
#SPJ4
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Your questions deserve reliable answers. Thanks for visiting IDNLearn.com, and see you again soon for more helpful information.