Calculating Net Electric Field at Origin: Two Charges on X-Axis, Two on Y-Axis

[thursdaytbs]

Electric Field question:

Two charges are located on the x axis: q1 = +6.0 micro-Coloumbs at x1=+4cm, and q2=+6 micro-coloumbs at x2 = -4cm. Two other charges are located on the y-axis: q3 = +3 micro-Coloumbs at y3=+5cm, and q4 = -8 micro coloumbs at y4 = +7cm. Find the net electric field (magnitude and direction) at the origin.

What I've tried to do is say E = F/q, and F = Kq1q2 / r^2, therefore E = (kq1q2 / r^2) / q. since q2 = origin, it = 1? and the q1 and q cancel out? So it becomes: E=k/r^2?

I'm pretty sure I'm doing it wrong and was wondering if someone could just point me in the right direction?

Any help appreciaited, thanks.

 

[dextercioby]

HINT:The electric (electrostatic) field created by a charge 'q' at the point \vec{r} is given by
\vec{E}_{q,\vec{r}}=:\frac{q}{4\pi\epsilon_{0}r^{2}}\frac{\vec{r}}{r}
,where \vec{r} is the position vector at the point u wish to calculate the fiels wrt to the point in which is the source "q".

Daniel.

PS.You'll have to apply that formula 4 times (for each charge) and then add those 4 vectors obtained.

 

[apchemstudent]

Electric Field question:

Two charges are located on the x axis: q1 = +6.0 micro-Coloumbs at x1=+4cm, and q2=+6 micro-coloumbs at x2 = -4cm. Two other charges are located on the y-axis: q3 = +3 micro-Coloumbs at y3=+5cm, and q4 = -8 micro coloumbs at y4 = +7cm. Find the net electric field (magnitude and direction) at the origin.

What I've tried to do is say E = F/q, and F = Kq1q2 / r^2, therefore E = (kq1q2 / r^2) / q. since q2 = origin, it = 1? and the q1 and q cancel out? So it becomes: E=k/r^2?

I'm pretty sure I'm doing it wrong and was wondering if someone could just point me in the right direction?

Any help appreciaited, thanks.

E = F/q(o) where q(o) is the charge experiencing the force.

so E = Kq1q(o)/(r^2*q(o)) <----q1 is the charge providing the force and the q(o) is the charge that would be at the origin. However though, as you mentioned before, it does cancel out.

 

[thursdaytbs]

Thanks for the help everyone.