Question

A 7.87 nC is located 1.91m from 3.90nC point charge.Find the magnitude of the electrostatic force that one charge exerts on the other.
step1:Fe=Ke=|q1|*|q2|/r^2
step2:(8.99x10^9)(7.787*3.90/1.912)=7.563665196e10
step3:correct answer is 7.56e-08
Can someone help me my answer was 7.563665196e10 when the right answer is 7.56e-08 what did i do wrong ?

Answer 1

\[\vec {\mathbf F}_e=K\frac{q_1 q_2}{r^2}\hat{\mathbf r}_{12}\]\[F_e=K\frac{q_1 q_2}{r^2}\]\[=9\times10^9[\text N \cdot\text m^2\cdot\text C^{-2}]\frac{7.87 [\text{nC}]\times 3.90[\text{nC}]}{(1.91[\text m])^2}\]\[=9\times10^9[\text N \cdot\text m^2\cdot\text C^{-2}]\frac{7.87\times10^{-9} [\text{C}]\times 3.90\times10^{-9}[\text{C}]}{(1.91[\text m])^2}\]

Answer 2

\[=\frac{9\times7.87\times 3.90}{(1.91)^2}\times10^{9-9-9}\frac{[\text N \cdot\text m^2\cdot\text C^{-2}][\text{C}][\text{C}]}{[\text m^2]}\]\[=\frac{9\times7.87\times 3.90}{1.91\times1.91}\times10^{-9}[\text N ]\]\[=75.6\times10^{-9}[\text N ]\]\[=7.56\times10^{-8}[\text N ]\]

Answer 3

i think your main error was forgetting that \([\text {nC}]=10^{-9}[\text C]\)

Answer 4

also you've mistyped the working in step two (the charge is slightly off and the square on the denominator is missing)

Answer 5

thank you=]

Answer 6

\[\color{goldenrod}{\ddot\smile}\]