Question

cakulate the coulombian force blw a proton
and an electron seperated by 0.8 x 10^15 m?​

Answer 1

Answer:

We know

F = $\frac{kq1q2}{r^{2} }$ ---------( 1 )

It is given to us that:

q1 = 1.6 x 10 ⁻¹⁹ C (Charge on Proton)

q2 = - 1.6 x 10 ⁻¹⁹ C (Charge on Electron)

r = 0.8 x 10¹⁵ m

k = 9 x 10⁹ Nm²C⁻²

∴ inserting in eqn (1)

=   $\frac{9*10^9 * (1.6* 10^-^1^9)^2}{(0.8*10^1^5)^2}$

∴ F = - 3.6 x 10⁻⁶⁸ N

The negative sign indicates the force is attractive in nature...