Question

If the electrical force is 4.0 Coulombs, while one point has a charge of 1 x 10 -6 C and the other point has a charge of 2 x 10 -6 C, what is the distance that separates them?

Answer 1

can u help?

Answer 2

\[F=\frac{ kq _{1}q _{2} }{r ^{2}}\] in your problem I want to say\[F=4, q _{1}=1x10^{-6},q _{2}=2x10^{-6}, k \approx9x10^{9}\]

Answer 3

can u explain?

Answer 4

F is the Force, K is coulomb's constat, q1 is the first charge, q2 is the second charge, r is the distance between the two atoms

Answer 5

ok

Answer 6

so how do i find the distance

Answer 7

\[so F = 9 x 10^9 1x 10^-6/\]

Answer 8

wait 2 x 10^-6

Answer 9

so i'll multiply 4 with the two charges right or?

Answer 10

????

Answer 11

plugging your knowns into the equation;
\[F=\frac{ kq _{1}q _{2} }{ r ^{2} }\] becomes\[4=\frac{ 9x10^{9}x10^{-6}x2x10^{-6} }{ r ^{2} }\] that becomes
\[r ^{2}=\frac{ 9x10^{9}x10^{-6}x2x10^{-6} }{ 4}\]
\[r ^{2}=0.0045\]
r=0.067082...