IDNLearn.com: Where your questions meet expert advice and community insights. Our Q&A platform offers reliable and thorough answers to help you make informed decisions quickly and easily.

Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 0,92 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles qı and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.500 m. What is net force on q3?


Sagot :

Answer:

1.12

Explanation:

it is correct on Acellus

The total force on charge q(3) is 1125 x [tex]10^{-3}[/tex] Newton

We have three charges in a straight line.

We have to find the force on charge q3 due to charge q1 and q2.

State Coulomb's law.

The force between two point charges is directly proportional to the product of the magnitude of two point charges and inversely proportional to the square of the distance between them. Mathematically-

[tex]F = \frac{1}{4\pi\epsilon_{o}}{\frac{qQ}{r^{2} } } }[/tex]

We have -

[tex]q_{1} =q_{2} =q_{3} = -5 \times 10^{-6} C[/tex]

The distance between -

[tex]q_{1}\;and\;q_{2} = q_{2} \;and\;q{3}[/tex] = 0.5 m

Now, Force on charge q(3) due to charge q(1) -

F(1 → 3) = [tex]\frac{1}{4\pi\epsilon_{o}}{\frac{5\times 10^{-6} \times 5\times 10^{-6} }{1^{2} } } } = 9\times 10^{9} \times 25 \times 10^{-12} = 225\times 10^{-3}[/tex] N (repelling force)

Force on charge q(3) due to charge q(2) can be calculate similarly as -

F(2 → 3) = 900 x [tex]10^{-3}[/tex] N(repelling force)

Now, both the forces are in same direction, so we can add them algebraically -

[tex]F_{net}[/tex] = F(1 → 3) + F(2 → 3) = 1125 x [tex]10^{-3}[/tex] N

Hence, the total force on charge q(3) is 1125 x [tex]10^{-3}[/tex] Newton

Hence, the wavelength of light in nano meters is 0.0177 x   nm.

To solve more questions on Coulomb's law, visit the link below -

https://brainly.com/question/14110379

#SPJ2

We are delighted to have you as part of our community. Keep asking, answering, and sharing your insights. Together, we can create a valuable knowledge resource. IDNLearn.com is your go-to source for accurate answers. Thanks for stopping by, and come back for more helpful information.