Question

Calculate the force of electrostatic attraction between
a proton and an electron separated by a distance
of 8 X 10-14 m.​

Answer 1

given that–

magnitude of charge on a proton (q1) = 1.6 x 10^-19 C

magnitude of charge on an electon (q2) = 1.6 x 10^-19 C

distance between them (r) = 8 x 10^-14 m

we know that,

$f_{electrostatic} \\ = \frac{k q_{1} q_{2} }{ {r}^{2} } = \frac{(9 \times {10}^{9})(1.6 \times {10}^{ - 19} )(1.6 \times {10}^{ - 19} )}{ {(8 \times {10}^{ - 14}) }^{2} } \\ = \frac{(9 \times {10}^{9})(2.56 \times {10}^{ - 38}) }{64 \times {10}^{ - 28} } \\ = \frac{(9 \times {10}^{9})(256 \times {10}^{ - 40} ) }{64 \times {10}^{ - 28} } \\ = \frac{9 \times 256 \times {10}^{9 - 40} \times {10}^{28} }{64} \\ = 9 \times 4 \times {10 }^{9 - 40 + 28} \\ = 36 \times {10}^{ - 3} \\ = 0.036 \: newton$

Hence, electrostatic force between the given proton and neutron is 0.036 N.

Answer 2

Answer:

your answer is

Explanation:

given that–

magnitude of charge on a proton (q1) = 1.6 x 10^-19 C

magnitude of charge on an electon (q2) = 1.6 x 10^-19 C

distance between them (r) = 8 x 10^-14 m

we know that,

\begin{lgathered}f_{electrostatic} \\ = \frac{k q_{1} q_{2} }{ {r}^{2} } = \frac{(9 \times {10}^{9})(1.6 \times {10}^{ - 19} )(1.6 \times {10}^{ - 19} )}{ {(8 \times {10}^{ - 14}) }^{2} } \\ = \frac{(9 \times {10}^{9})(2.56 \times {10}^{ - 38}) }{64 \times {10}^{ - 28} } \\ = \frac{(9 \times {10}^{9})(256 \times {10}^{ - 40} ) }{64 \times {10}^{ - 28} } \\ = \frac{9 \times 256 \times {10}^{9 - 40} \times {10}^{28} }{64} \\ = 9 \times 4 \times {10 }^{9 - 40 + 28} \\ = 36 \times {10}^{ - 3} \\ = 0.036 \: newton\end{lgathered}felectrostatic=r2kq1q2=(8×10−14)2(9×109)(1.6×10−19)(1.6×10−19)=64×10−28(9×109)(2.56×10−38)=64×10−28(9×109)(256×10−40)=649×256×109−40×1028=9×4×109−40+28=36×10−3=0.036newton

Hence, electrostatic force between the given proton and neutron is 0.036 N.