1.

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

Answer»

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.


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