Question

An astronaut on the ISS is doing an experiment with two puffy Cheetos inside a carefully sealed container. A charge of 1.2 E−5 C is place on the first Cheeto. The second one receives a charge of 2.5 E−6 C. The force acting on the first Cheeto is 5.73 E−1 N. What is the distance between the Cheetos if k = 8.99 E9 N*m2/C2?70 cm60 cm40 cm49 cm

Answer 1

1st option: 70 cm

Explanation:

Given:

Force (F) = 5.73 E−1 N^

Electrical charge 1 (q1) = 1.2 E−5 C

Electrical charge 2 (q2) = 2.5 E−6 C

k = 8.99 E9 N*m2/C2

We can calculate the distance between both charges with the help of Coulumb's law, just like this:

`F=k\cdot(q_1\cdot q_1)/(d^2)`

We substitute and solve for the distance, just like this:

`\begin{gathered} 5.73\cdot\: 10^(-1)=(8.99\cdot10^9\cdot\:1.2\cdot\:10^(-5)\cdot2.5\cdot10^(-6))/(d^2) \\ d^2=(8.99\cdot10^9\cdot1.2\cdot10^(-5)\cdot2.5\cdot10^(-6))/(5.73\cdot10^(-1)) \\ d=\sqrt[]{0.47} \\ d=0.69\text{ m}\cong0.70\text{ m}=70cm \end{gathered}`

The distance is 70 centimeters