Question

Calculate the distance between proton and electron in air if the attract each other with a force of 6*10^-11 Newton plz answer fast

Answer 1

Answer:

$\boxed{\mathfrak{Distance \ between \ proton \ and \ electron = 2 \ nm}}$

Given:

Force of attraction (F) between proton & electron = $\sf 6 \times 10^{-11}$ N

To Find:

Distance (r) between proton & electron

We know:

Magnitude of charge on proton ($\sf q_p$) = $\sf 1.6 \times 10^{-19}$ C

Magnitude of charge on electron ($\sf q_e$) = $\sf 1.6 \times 10^{-19}$ C

$\sf k = \dfrac{1}{4\pi \epsilon_0} = 9 \times 10^9 \ Nm^2/C^2$

Explanation:

From Coulomb's inverse square law:

$\boxed{ \bold{F = \dfrac{kq_p q_e}{r^2}}}$

By substituting value of F, k, $\sf q_p$ & $\sf q_p$ in the equation we get:

$\sf \implies 6 \times {10}^{ - 11} = \dfrac{9 \times {10}^{9} \times 1.6 \times {10}^{ - 19} \times 1.6 \times {10}^{ - 19} }{ {r}^{2} } \\ \\ \sf \implies 6 \times {10}^{ - 11} = \dfrac{9 \times {10}^{9} \times {(1.6 \times {10}^{ - 19}) }^{2} } { {r}^{2} } \\ \\ \sf \implies 6 \times {10}^{ - 11} = \dfrac{9 \times {10}^{9} \times 2.56 \times {10}^{ - 38} }{ {r}^{2} } \\ \\ \sf \implies 6 \times {10}^{ - 11} = \dfrac{23.04 \times {10}^{ - 38 + 9} }{ {r}^{2} } \\ \\ \sf \implies 6 \times {10}^{ - 11} = \dfrac{23.04 \times {10}^{ - 29} }{ {r}^{2} } \\ \\ \sf \implies {r}^{2} = \dfrac{23.04 \times {10}^{ - 29} }{6 \times {10}^{ - 11} } \\ \\ \sf \implies {r}^{2} = 3.84 \times {10}^{ - 29 - ( - 11)} \\ \\ \sf \implies {r}^{2} = 3.84 \times {10}^{ - 29 + 11} \\ \\ \sf \implies {r}^{2} = 3.84 \times {10}^{ - 18} \\ \\ \sf \implies r = \sqrt{3.84 \times {10}^{ - 18}} \\ \\ \sf \implies r \approx 2 \times {10}^{ - 9} \: m\\ \\ \sf \implies r \approx 2 \: nm$

$\therefore$

Distance (r) between proton & electron = 2 nm

Answer 2

Answer:

F = 6 × 10^-11 N

q = 1.6 × 10^-19 C

k = 9 × 10^9 Nm²/C²

F = kq²/r²

By substituting value we get:

r = 2 × 10^-9 m