Multi-Select Question

A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi

zzz [600]

Answer:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Explanation:

We know that the electric force equation is:

$F=k\frac{q_{1}*q_{2}}{r^{2}}$

k is the electric constant $9*10^{9} Nm^{2}/C^{2}$
r is the distance between the particles
q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

$F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}$

$F_{31}=k\frac{q*2q}{r_{31}^{2}}$

$F_{31}=k\frac{2q^{2}}{r_{31}^{2}}$

r(31) is the distance between 3 and 1

2. Now, let's find the electric force between the third particle and the second particle.

$F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}$

$F_{32}=k\frac{q*(-q)}{r_{32}^{2}}$

$F_{32}=-k\frac{q^{2}}{r_{32}^{2}}$

r(32) is the distance between 3 and 2.

Now, $r_{31}+r_{32}=2a$ or $r_{32}=2a-r_{31}$

The net force must be zero so:

$F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}$

We just need to solve it for r(31)

$r_{31}^{2}=2(2a-r_{31})^{2}$

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Therefore the distance from the origin will be:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

I hope it helps you!

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4 years ago

Write a scientific explanation to describe where earthquakes and volcanoes most commonly occur and their relation to plate bound

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Earthquakes and volcanoes most commonly occur around plate boundaries because of the movement from the plate boundaries. The interactions between the plates by moving under, upon, or sliding against other boundaries may cause earthquakes and volcanoes.

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Rays of light incident on a lens and parallel to the principal axis of the lens coverage?

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B at the focal point of the lens

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4 years ago

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Find out the x-component and y-component of the following vectors:

Leno4ka [110]

Answer:

A = 15.0 cm; the vector is 30.00 north of west; = 15 cm N30W

Y - component = 7.5cm and X - component = -12.99 cm

B = 12.0 m; the vector is 55.00 north of east; = 12.0 m N55E

Y - component = 9.83 m and X - component = 6.88 m

C = 28.0 m/s; the vector is 40.00 south of west; = 28.0 m/s S40W

Y - component = - 17.998 m/s and X - component = - 21.448 m/s

D = 55.5 m/s2; the vector is 35.00 south of east; = 55.5 m/s² S35E

Y - component = - 31.835 m/s² and X - component = 45.466 m/s²

Explanation:

1. A = 15.0 cm; the vector is 30.00 north of west; = 15 cm N30W

Y - component = 15cm × sin30 = 7.5cm

X - component = -15cm × cos30 = -12.99 cm

2. B = 12.0 m; the vector is 55.00 north of east; = 12.0 m N55E

Y - component = 12.0 m × sin55 = 9.83 m

X - component =12.0 m × cos55 = 6.88 m

3. C = 28.0 m/s; the vector is 40.00 south of west; = 28.0 m/s S40W

Y - component = - 28.0 m/s × sin40 = - 17.998 m/s

X - component = - 28.0 m/s × cos40 = - 21.448 m/s

4. D = 55.5 m/s2; the vector is 35.00 south of east; = 55.5 m/s² S35E

Y - component = - 55.5 m/s² × sin35 = - 31.835 m/s²

X - component = 55.5 m/s² × cos35 = 45.466 m/s²

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3 years ago

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What happens when sound waves move faster?

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Hnjjjjj hejehtjriwje a wkirrhnemekwhr. Rthejb wine rnrie

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