3 point charges are located on the corners of an equilateral triangle. the sides of which being 25 cm in length. the charges are 4.50 x 10^-6 C, 2.50 x 10^-6 C, and -4.00 x 10^-6 C. calculate net electrostatic charge on the 2.50 x 10^-6 C charge.
2007-03-20 10:08:09 · 2 answers · asked by avemaria 2 in Science & Mathematics ➔ Physics
Here's my best guess at it. [edit: I totally misread the question, my bad. anyway].
Net electrostatic charge = Electric intensity on the charge right? [Well I think it is]
Electric intensity = E= F/q
F = Force experienced by the charge, q = test charge.
F = 1/4πεо x Q [q]/r^2
Where εо = permitivity of free space = 8.85 x 10^12
[Q] test charge = 2.50 x 10 ^-6
q1= point charge 1 = 4.50 x 10^-6
q2 = point charge 2 = -4.00 x 10^-6
r^2= square of distance = 25 x 25 = 625 cm
To calculate the force exerted by q1 on Q, subsitute the value of q1 for [q] in the eq.
To calculate the value of q2 substitute the value of q2 for [q] int he eq.
Add the two answers up to get the net force. Then divide by Q.
I hope I'm right. Good luck.
2007-03-20 11:04:44 · answer #1 · answered by sarahh_f 3 · 0⤊ 0⤋
F=(kq1q2)/(r^2)
k=8.98755*10^9
F1=(k(4.5*10^6)(2.5*10^6))/.25^2 away from the charge the charge
F2=(k(4.0*10^6)(2.5*10^6))/.25^2 towards the charge
split the forces into x and y components
add the x and y components together then find the total force
it ends up being 1.5358Newtons
2007-03-20 18:27:13 · answer #2 · answered by jcj7373 2 · 0⤊ 0⤋