Question

electric force is acting between two electrons placed at certain distance. What will be the force between two protons separated by half of the distance of electron


I need explanation ​

Answer 1

AnswEr :

Suppose two electrons are placed at a distance 'r' , then the protons will be placed at a distance (r/2)

We know that,

Magnitude of charge of an electron and a proton are equal .

Using the Formula,

$\sf \: F = \dfrac{1}{4 \pi \epsilon_o} \times \dfrac{Qq}{r {}^{2} }$

$\mathbb{HERE}\begin{cases}\sf{F \longrightarrow Force} \\ \sf{\epsilon_o \longrightarrow Permittivity \ of \ Free \ Space} \\ \sf{Q,q \longrightarrow Charges } \\ \sf{r \longrightarrow Separation \ Distance } \end{cases}$

Let $\sf F_1$ be the force acting between two electrons

$\sf \: F_1 = \dfrac{1}{4\pi \epsilon_o} \times \dfrac{q_e {}^{2} }{ {r}^{2} } - - - - - - - (1)$

Let $\sf F_2$ be the force acting between two protons

$\sf \: F_2 = \dfrac{1}{4\pi \epsilon_o} \times \dfrac{ {q_p}^{2} }{( \frac{r}{2}) {}^{2} } \\ \\ \longrightarrow \: \sf \: F_2 = \dfrac{1}{4\pi \epsilon_o} \times \dfrac{4 {q_p}^{2} }{r {}^{2} } - - - - - (2)$

Dividing equations (1) and (2),we get :

$\implies \sf \dfrac{F_1}{F_2} = \cancel{ \dfrac{4\pi \epsilon_o}{4\pi \epsilon_o} } \times \dfrac{4 \cancel{ {q_e}^{2} {r}^{2}} }{ \cancel{q_p{}^{2}r {}^{2} } } \\ \\ \implies \: \boxed{ \boxed{ \sf \: F_2 = \dfrac{F_1}{4}}}$