91Ó°ÊÓ

The given data can be listed below,

• The charge on the proton is$q_p$.
• The electric field due toproton on proton at location B is${\stackrel{→}{E}}_1$,
• The force experience by another proton at location B is, $\stackrel{→}{F}$.
• Lithium nucleus placed in place of proton at location B.
• The charge on the lithium nucleus is,$q=3q_p$

The coulomb force (also known as electrostatic force) is proportional to the product of two "point-like" electric charges and inversely proportional to the square of their separation. Because charge can be either positive or negative, the result can be either positive (repulsive force) or negative (attractive force).

The electric field can be determined by using the electric force and charge on that point A. So, the electric filed on point B will depend on the charge of proton on point A even after the replacement of particle at point B.

Thus, the electric field at point B due to proton is${\stackrel{→}{E}}_1$

The force on the lithium nucleus is given by,

${\stackrel{→}{F}}_L={\stackrel{→}{E}}_{1}q$

Here, ${\stackrel{→}{E}}_1$is the electric field and q is the charge on lithium nucleus.

Substitute values in the above,

$$\begin{gathered} {\stackrel{→}{F}}_L=3{\stackrel{→}{E}}_1{q}_p \\ =3{\stackrel{→}{F}}_1 \end{gathered}$$

Thus, the force on lithium nucleus is$3{\stackrel{→}{F}}_1$ .

The electric field at location/point B is due to the charge at location A. so when the electron is placed at location B at the place of nucleus the electric field on location B will not be affected by it.

Thus, the electric filed at location B is ${\stackrel{→}{E}}_1$due to the proton at location A.

The magnitude of the force is calculated by taking the magnitude of charge $q=1.6×{10}^{-19}C$that is the charge on proton.so, electron have same force as on proton but opposite in direction.

The magnitude of force on the electron is given by,

$$\begin{gathered} \stackrel{→}{F}=\stackrel{→}{E}q \\ =-{\stackrel{→}{F}}_1 \end{gathered}$$

Here, E is the electric field and q is the charge on the electron i.e., $1.6×{10}^{-19}C$.

Thus, the magnitude of force on electron is $-{\stackrel{→}{F}}_1$.

The direction of force exerted on electron is opposite to the force exerted on the proton as shown in the figure the force acting on the proton is in the direction of arrow b. so the direction of force for electron must be shown by arrow f.

Thus, the arrow f indicates the direction of force on electron due to the electric field.