Physics

Physics, 30.08.2019 16:10, barstr9146

Two equal positive charges q1 = q2 = 2μc are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. what are the magnitude and direction of the total electric force that qı and q2 exert on a third charge q = -4 μc at x = 0.40 m, y = 0.15 m?

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Answer

answered: rprest00

$Fx=486.58(-i)*10^{-2} N$ N: Resulting force on Q in the direction of the x axis

$Fy=162.49j(j)*10^{-2}$ N: Resulting force on Q in the direction of the y axis

Explanation:

To solve this problem we need 3 basic concepts:

1.Coulomb's law:

Two point charges (q1, q2) separated by a distance (r) exert a mutual force (F) whose magnitude is determined by the following formula:

$F=k*(\frac{q1*q2}{r^{2} } )$ Formula (1)

$K=8.99*10^{9} \frac{N*m^{2} }{C^{2} }$ : Coulomb constant

q 1,q2=charge in Coulombs (C)

The direction of F is determined by the unit vector ( $r_{u}$ ) in the direction of F

$r=r_{x} +r_{y}$ : position vector in the direction of F

r= (x2-x1)i+( y2-y1)j

The points (x1, y1) and (x2, y2) represent the position coordinates of the charges .

The sense of force depends on the sign of q1 and q2:

When the charges have opposite signs the force is attractive.

When the charges have an equal sign, the force is repulsion

2. Newton's third law: law of action and reaction

The magnitude of the force exerted by q1 on q2 is equal to the force exerted by q2 on q1, and these forces have the opposite direction.

F1-2=-F2-1

Problem development

We have the forces exerted by the charges q1 + and q2 + on the load Q-.

The forces F1 and F2 on the load Q are attractive forces, Because of this the force comes out of Q and is directed towards q1 and q2

Calculation of vector r1 : points Q(0.4,0.15) , q1(0,0.3)

r1= (0- 0.4)i+(0.3-1.5)j

r1=-0.4i+0.15j

magnitude of r1

$r1=\sqrt{-0.4^{2}+0.15^{2} } =\sqrt{0.16+0.0225} =0.1825m$

Calculation of vector r2 : points(0.4,0.15) (0,-0.3)

r2= (0- 0.4)i+(-0.3-1.5)j

r2=(-0.4i-0.45j) m

magnitude of r2

$r2=\sqrt{-0.4^{2}+(-0.45)^{2} } =\sqrt{0.16+0.2025} =0.6 m$

We apply formula 1 to calculate the magnitudes of F1 and F2 with the known data :

q1 = q2 = 2μC , r1= 0.1825m ,r2=0.6m,

$K=8.99*10^{9} \frac{N*m^{2} }{C^{2} }$

1 μC= $10^{-6} C$

$F1=k*(2*10^{-6} *4*10^{-6} )/(0.1825^{2} )$

$F1=k*240.195*10^{-12}$

$F1=8.99*10^{9} *240.195*10^{-12}=2159.352*10^{-3} =215.93*10^{-2} N$

$F2=k*(2*10^{-6}*4*10^{-6}) /(0.6)^{2}$

$F2=8.99*10^{9} *22.2*10^{-12} =199.77*10^{-3} =19,97*10^{-2} N$

Calculation of the total electric force (F) exerted by the charges q1 and q2 on the charge Q

$r_{u}=(r_{x} +r_{y} )/r$ : unit vector in the direction of r

$F=F1(r_{u1})+F2(r_{u2} )$

ru=(rx+ry)/r

$F=\frac{215.93*(-0.4i+0.15j)}{0.1825} +\frac{19.93*(-0.4i-0.45j}{0.6} *10^{-2} N$

F=(-473.27i+177.47j-13.31i-14.98j) ))*10e-2 N

$F=(-486.58i+162.49j)*10^{-2} N$

$Fx=486.58(-i)*10^{-2} N$ N: Resulting force on Q in the direction of the x axis

$Fy=162.49j(j)*10^{-2}$ N: Resulting force on Q in the direction of the y axis

Answer

answered: Guest

Answer: fn = wcos50 = 38.89(9.81)(cos50) = 245.23 n

Answer

answered: Guest

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