Saturday

August 30, 2014

August 30, 2014

Posted by **AfterLife** on Sunday, August 28, 2011 at 9:52pm.

-- I am trying to post a graph but it is not willing to so I will try to explain it --

Q+ is up on the y access ( x = 0 ) and Q- is down the y access ( x = 0 ) , both at equal distances d of the origin. q is to the right of the x access ( y = 0 ) at a distance x from the origin

So I am trying to solve it like this

Net Force = (Force of +Q on q) + ( Force of -Q on q)

using this equation for force (Kq1q2)/r^2

but then we need to find the components right ?

so I thought it is like this

Fnet = sqrt ( Fnet of x components )^2 + (Fnet of y component)^2

but then for the x components , the r in the equation should be x ,, and for the y components the r should be d

am I right ? I feel confused and I think I am wrong :S

- Physics -
**bobpursley**, Sunday, August 28, 2011 at 10:11pmYou are almost right, however, it is easier for me to break up the two forces into vertical and horizontal components.

well, first, the force on q from Q is kQq/distance^2 at some angle from the horizonal (arc tan d/x)

the horizontal force is

a) kqq/distanc^2*cosTheta

if you calculate the horizontal force component from -Q, it is the same but in a opposite direction, so the net horizontal force is ZERO.

Vertical forces:

vertical force from Q (top) is

kQq/distance^2 * sinTheta in the down direction

vertical force from -Q (bottom) is the same, and in the down direction.

so the net force is downward,

2kQq/(d^2+x^2) * sin (arctan d/x)

- Physics -
**AfterLife**, Sunday, August 28, 2011 at 10:12pmThank you very much :D

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