Asked by AfterLife
Determine the magnitude and direction of the total electric force on the +q charge by the +Q and -Q charges respectively. it should express and algebraic expression using Q,q,x,d and fundamental constants.
-- 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
-- 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
Answers
Answered by
bobpursley
You 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)
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)
Answered by
AfterLife
Thank you very much :D
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