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?

Answers: 3

Show answers

Answers

Answer from: 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 from: Quest

The answer is c hope this

Answer from: Quest

Ihope you find the answer

Another question on Physics

Physics, 21.06.2019 19:10

What happens to light waves from a star as the star moves away from earth?

Answers: 3

Answer

Physics, 22.06.2019 05:30

Mutt and jeff are running along a path each pushing a cart loaded with rocks. if mutt's wagon has a mass two times greater than jeff's and they want to stay together, arriving at their destination at the same time, what must happen? a) jeff should slow down so mutt can catch up. b) mutt must use half as much force to push his cart. c) mutt must use twice as much force to push his cart. d) mutt and jeff will use the same force, but mutt will take bigger steps.

Answers: 1

Answer

Physics, 22.06.2019 06:50

Much of the information taught in earth-space science would not be available if it weren’t for our remote sensing abilities. explain two ways our remote sensing strategies contribute to our knowledge of these topics.

Answers: 3

Answer

Physics, 22.06.2019 15:00

What happens to an isotope during radioactive decay?

Answers: 2

Answer

You know the right answer?

Two equal positive charges q1 = q2 = 2μc are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, re...

Questions