If the force between two equal but opposite charges is 0.5 N, when they are 20 cm apart, calculate the magnitude of the charges.
2006-12-11 20:48:40 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
som almost had the right answer but he forgot to divide the whole thing by 4pi
so the eqation that you have to use is
f = Q^2/(4*pi*Eo*d^2)..
where
q is the magnitude of the charges
Eo = electrical permitivity = 8.85*10^-12
and d is the distance between the charges this has to be in meters. only then will you get the right answer.
on more thing that helps is
1/4*pi*Eo = 9*10^9
all u have to do now is plug in the values and get the answer.
dont worry about sign on the charges.
2006-12-11 21:09:21 · answer #1 · answered by zero 2 · 0⤊ 0⤋
Let the charge be Q Coulombs
F=Q^2/(E0*d)
E0=8.85*10^-12
Q^2=0.5*0.2*8.85*10^(-12)
Q =9.40966*10^(-7) C
Q =94.1 microcoulomb
2006-12-12 04:54:59 · answer #2 · answered by Som™ 6 · 0⤊ 0⤋
I am sorry I am poor in calculation. Why dont you approch a physics professor.
2006-12-12 04:53:33 · answer #3 · answered by Kumari V 3 · 0⤊ 0⤋