KEAM 2021 Physics and Chemistry Solved Paper

Electrostatic force between two charges q1 and q2 is given as
F=‌

Kq1q2

r2

where, K is the Coulomb's constant and r is the distance between the charges.
According to the question, for two protons with charge q (say) and at a distance r apart, the electrostatic force is given as
Fp=‌

Kqq

r2

=‌

Kq2

r2

.....(i)
Similarly, for two electrons with charge e and at a distance r apart, the electrostatic force is given as
Fe=‌

Kee

r2

=‌

Ke2

r2

.....(ii)
On dividing Eq. (i) by Eq. (ii), we get
‌

Fp

Fe

=‌

Kq2

r2

/‌

Ke2

r2

⇒ ‌

Fp

Fp

=‌

q2

e2

Since, the magnitude of charge on proton = the magnitude of charge on electron i.e.
|q|=|e|=1.6×10−19C
⇒ |‌

Fp

Fe

|=|‌

q2

e2

|=|‌

q2

q2

|=‌

1

1

Or |Fp:Fe|=1:1