Question

done

Answer 1

can someone help me figure out the equation

Answer 2

\(F_e = k\dfrac{q_1q_2}{d^2} \)

k is coulomb's constant: \(k = 8.99 \times 10^9 Nm^2C^{−2}\)

So it looks like this:

\(2 = 8.99 \times 10^9 \dfrac{q^2}{1.35^2} \), because the charges are the same. Solve for q

pretty sure Voca has already explained this

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html

Answer 3

@sillybilly123 he was offline and I had questions. Plus he closed my questions so I had to re-open it. Thanks anyways.

Answer 4

sure, no worries