Question

An electron and proton are placed in an electrical field. The forces acting on them are $\overrightarrow{F_1}$ and $\overrightarrow{F_2}$ and their accelerations are $\overrightarrow{a_1}$ and $\overrightarrow{a_2}$ respectively, then

• A. $\overrightarrow{F_1}=\overrightarrow{F_2}$
• B. $\overrightarrow{F_1}+\overrightarrow{F_2}$ $=0$
• C. $\mid \overrightarrow{a_1}\mid =\mid \overrightarrow{a_2}\mid$
• D. $\mid \overrightarrow{a_1}\mid \ge \mid \overrightarrow{a_2}\mid$

Answer 1

Answer: (B), (D)

The correct options are
B $\overrightarrow{F_1}+\overrightarrow{F_2}$ $=0$
D $\mid \overrightarrow{a_1}\mid \ge \mid \overrightarrow{a_2}\mid$

magnitude of charge on electron = magnitude of charge on proton

but they are opposite in sign. so force exerted upon them will be opposite in sign too.

Force = mass*accerlation

also,

$F=\frac{K{q}_1{q}_2}{r^2}=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{r^2}$

${ϵ}_0=8\cdot 85\times {10}^{-12}\frac{C^2}{N-m^2}=$ permittivity of the free space or vacu

um

so

force on electron = $-1\ast$force on proton

or $F_1+F_2=0$

Basically both of them will attract each other.

Also

mass of electron is less than mass of proton, So $a_1>a_2$