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Identical 50 μC charges are fixed on an x axis at x = ±3.0 m. A particle of charge q = -15 μC is then released from rest at a point on the positive part of the y axis. Due to the symmetry of the situation, the particle moves along the y axis and has kinetic energy 1.2 J as it passes through the point x = 0, y = 4.0 m.
(a) What is the kinetic energy of the particle as it passes through the origin?
(b) At what negative value of y will the particle momentarily stop?

Respuesta :

tatendagota

Answer:

14 J

Explanation:

Let the angle of charge be θ

Then the distance from the x-axis = 3 m

The angle is given as [tex]tan (\theta) = \frac{3}{3}\\ \theta = tan^{-1} (1)\\ = 45[/tex]

Similarly, the charge is on the left side, so the distance will be - 3 m

The angle therefore will be  = 45°

So we need to find the net force given by the law:

[tex]F = \frac{kq_{1}q_{2} }{r^{2} }[/tex]

The distance between the points will be [tex]\sqrt{(3)^{2} +(3)^{2} }[/tex]

= 3[tex]\sqrt{2}[/tex]

                                                           

(a) The kinetic energy of the particle as it passes through the origin is 3J.

(b) The particle will momentarily stop at y = ±8.48 m

Conservation of energy:

The given position of the block is x = 0, y = 4.0 m

So the distance r from each charge Q = 50 μC at x = ±3.0 m wii be:

r = [tex]\sqrt{x^2+y^2}[/tex]

[tex]r=\sqrt{(0-3)^2+(4-0)^3}\\\\r=5m[/tex]

So the total electrostatic potential energy of charge q = -15 μC due to both the charges will be:

[tex]PE = \frac{1}{4\pi \epsilon_o} \frac{2Qq}{r}\\\\PE=9\times10^9\times\frac{2\times50\times(-15)\times10^{-12}}{5}\\\\PE= -2.7 \;J[/tex]

So the total energy of the charge at r = 5m is:

U = KE + PE = 1.2 - 2.7

U = -1.5 J

Now the distance of the charge q from the charge Q when it is at the origin is r = 3m. So the electrostatic potential energy at origin is:

[tex]PE' = \frac{1}{4\pi \epsilon_o} \frac{2Qq}{r}\\\\PE'=9\times10^9\times\frac{2\times50\times(-15)\times10^{-12}}{3}\\\\PE'= -4.5 \;J[/tex]

Let the kinetic energy at the origin be KE'

From, the law of conservation of energy, the total mechanical energy must remain conserved. Therefore

U = KE' + PE'

-1.5 = KE' + (-4.5)

KE' = 3 J

(b) The kinetic energy of the particle will be zero at the point it momentarily stops, so:

U = PE = -1.5J

[tex]-1.5= \frac{1}{4\pi \epsilon_o} \frac{2Qq}{r}\\\\-1.5=9\times10^9\times\frac{2\times50\times(-15)\times10^{-12}}{r}\\\\r= 9m[/tex]

the x coordinate will be zero as it lies on the y-axis, so:

[tex]9=\sqrt{(0-3)^2+(y-0)^2}\\\\y^2=72\;m^2[/tex]

y = ±8.48 m

Learn more about conservation of energy:

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