Question

Two equal unlike charges placed 3 cm apart in air attract each other with a force of 40 N. The magnitude of each charge is​

Answer 1

Answer:

as we know....

F =k Q q/ r^2

40= k q^2/(0.03)^2

40= 9×10^9 q^2/(0.03)^2

after solving ....

we get q=2×10^-5

hope it works

plz mark it as brainliest answer

Answer 2

Answer:

So, the magnitude of each charge is 200μm

Explanation:

The fee of the electron is equal to the value of the basic fee ( e ) however bearing a terrible sign. Since the fee of basic fee ( e) is roughly =1. 6×10−19 Hence fee of electron is =−1. 6×10−19 so from right here we will see the Magnitude of digital fee is ∣e∣=1.

F=Kq²/r²

q²=40 x 3 x3 / 3 x 10^9  

q=20 x 10^-5

q=200μm

So, the value of every fee is 200μm

#SPJ3