Question

Calculate the magnitude of electrostatic force between a proton and an electron separated by a distance of 0.5A .Given that magnitude of charge of proton and electron to be 1.6×10^-19C each and 1/4πÈ°=9×10^9Nm2 C-2

Answer 1

Given:

Proton and electron is separated by 0.5 A°. Magnitude of charge in electron and proton is 1.6 × 10^(-19) C.

To find:

Electrostatic force between proton and electron.

Calculation:

Electrostatic Force be F ;

$\therefore \: F = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ \dfrac{(q1)(q2)}{ {d}^{2} } \bigg \}$

$= > F = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ \dfrac{(1.6 \times {10}^{ - 19} ) \times (1.6 \times {10}^{ - 19} )}{ {(0.5 \times {10}^{ - 10}) }^{2} } \bigg \}$

$= > F = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ \dfrac{2.56 \times {10}^{ - 38} }{ 0.25 \times {10}^{ - 20} } \bigg \}$

$= > F = \dfrac{1}{4\pi\epsilon_{0}} \bigg \{ 10.24 \times {10}^{ - 18} \bigg \}$

Putting value of Coulomb's Constant:

$= > F = 9 \times {10}^{9} \times \bigg \{ 10.24 \times {10}^{ - 18} \bigg \}$

$= > F = 92.16 \times {10}^{ - 9} \: N$

So final answer is:

$\boxed{ \bold{ \large{ F = 92.16 \times {10}^{ - 9} \: N }}}$

Answer 2

Answer:

Here is your answer........