Q. $\mathbf{F}_{pe}\ $represents electrical force on proton due to electron and $\mathbf{F}_{ep}$on electron due to proton in a hydrogen atom.Similarly $\mathbf{F}_{pe}\ $represents the gravitational force on proton due to electron and $\mathbf{F}_{ep}\ $the corresponding force on electron due to proton. Which of the following is not true?
A) $\mathbf{F}_{Pe}\ + \ \mathbf{F}_{ep} = \ 0$
B) $\mathbf{F}’_{Pe}\ + \ \mathbf{F}’_{ep} = \ 0$
C) $\mathbf{F}_{Pe}\ + \ \mathbf{F}’_{Pe}\ + \ \mathbf{F}_{ep}\ + \ \mathbf{F}’_{ep} = \ 0$
D) $\mathbf{F}_{Pe}\ + \ \mathbf{F}’_{Pe}\ = \ 0\ $
Explanation
(d) In vector form of Coulomb’s law proves that the forces $\mathbf{F}_{12}\ and\ \mathbf{F}_{21}\ $are equal and opposite.
or $\mathbf{F}_{21} = \mathbf{F}_{12}$
$\mathbf{\ \ \ \ \ \ \ \ \ \ \ \ }{\mathbf{\ }\mathbf{F}}_{pe} = \ \mathbf{F}_{ep}$
$\mathbf{F}’_{pe} = \ \mathbf{F}’_{ep}$
$And\ \ \ \ \mathbf{F}_{pe} + \ \mathbf{F}_{ep} = \ - \ \mathbf{F}’_{ep}\ + \ \mathbf{F}’_{pe}$
So option (d) is incorrect.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.