Question

Two electron separated by a distance $r$ experiences force $F$ between them . Then force between a proton and a singly ionized helium atom separated by a distance $2r$ is

• A. $4F$
• B. $2F$
• C. $\frac{F}{2}$
• D. $\frac{F}{4}$

Answer 1

The correct option is D $\frac{F}{4}$

According to Coulomb's law

$\begin{array}{c}F=\frac{1}{4\pi E_0}.\frac{\left(e\right)\left(e\right)}{r^2}\\ F_1=\frac{1}{4\pi E_0}.\frac{\left(e\right)\left(e\right)}{r^2}\end{array}$

Charge on proton is equal to charge on electron and charge on singly Ionized

atom separated by distance ${}^{\prime }2r^{\prime }$

$\begin{array}{c}{F}_2=\frac{1}{4\pi E_0}.\frac{\left(e\lambda c\right)}{{\left(2r\right)}^2}\\ =\frac{1}{4\pi E_0}.\frac{e^2}{4r^2}\\ \frac{F_1}{F_2}=\frac{F}{4}\end{array}$

Option D.