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ziro4ka [17]
3 years ago
15

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 mag

nitude 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?
Physics
1 answer:
creativ13 [48]3 years ago
8 0

Answer:

Fx=486.58(-i)*10^{-2} NN: 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.

<h3>F1-2=-F2-1 </h3>

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

Answer:

Fx=486.58(-i)*10^{-2} NN: 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

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In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

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While bats emit a wide variety of sounds, one type emits pulses of sound at a frequency between 39 kHz and 78 kHz. What is the r
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A bat emits pulses of sound at a frequency between 39 kHz and 78 kHz. It is required to find the range of wavelengths of this sound.

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Free point cuz i cewl like dat‍♀️‍♂️
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Yes thank u teehee




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