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krishawnnn
14.09.2019 •
Physics
What is the net electrostatic force (magnitude and direction) on a particle with charge +5μc situated at the apex of an equili! ateral triangle if each of the other corners contain identical charges of-6 μc and the length of a side of the triangle is 0.10 m?
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Ответ:
net electrostatic force = 46.76 N
and direction is vertical downward
Explanation:
given data
charge = +5μC
corners identical charges = -6 μC
length of side = 0.10 m
to find out
What is the net electrostatic force
solution
here apex is the vertex where the two sides of equal length meet
so upper point A have charge +5μC and lower both point B and C have charge -6 μC
so
force between +5μC and -6 μC is express as
force =
..............1
put here electrostatic constant k = 9 ×
Nm²/C² and q1 q2 is charge given and r is distance 0.10 m
so
force =![9*10^{9} \frac{5*6*10^(-12)}{0.10^2}](/tpl/images/0231/3568/34cf7.png)
force = 27 N
so net force is vector addition of both force
force =![\sqrt{x^{2}+x^{2}+2x^{2}cos60}](/tpl/images/0231/3568/c12ef.png)
here x is force 27 N
force =![\sqrt{27^{2}+27^{2}+2(27)^{2}cos60}](/tpl/images/0231/3568/e2441.png)
net electrostatic force = 46.76 N
and direction is vertical downward
Ответ: