Question

electrostatic force between two positive ions carrying equal charges is 4×10N when separation is 5 Angstrom how many electrons are missing from each

Answer 1

Using Coulomb,'s law

F=Kq1q2/r2=Kq2/r2F=Kq1q2/r2=Kq2/r2

You get

q=r×√(F/K)q=r×√(F/K)

=(5×10−10)×√(4×10−9)/(9×109)=(5×10−10)×√(4×10−9)/(9×109)

= 3.33×10−193.33×10−19

Which is impossible as charge is quantized. It can have values which are integral multiple of 1.6 × 10^-19 only.

Second answer



Let's assume the charges to be +q.

F = kq2r2kq2r2

k= 9×109Nm2C−29×109Nm2C−2,

F=4×10−9N4×10−9N,

r=5×10−10m=5×10−10m

4×10−9 = 9×109.q225×10−20⟹4×10−9 = 
9×109.q225×10−20

Or,
q = 3.3×10−19C3.3×10−19C

q = -ne

⟹n=−3.3×10−19−1.6×10−19⟹n=−3.3×10−19−1.6×10−19

No. Of electrons missing in each, n= 2