Question

calculate the Magnitude of electrostatic force between a proton and an electrom separated by a distance 0.5 A

Answer 1

Explanation:

F=k.q1.q2/R^2

given q1=q2=1.6×10^-19

k=9×10^9

R=0.5×10^-10m

F=9×10^9×(1.6×10^-19)^2/(5×10^-11)^2

F=4.608×10^-7

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Answer 2

Question :-

Calculate the magnitude of electrostatic force between a proton and an electron seperated by a distance of 0.5 Å .

Answer :-

$\leadsto \red{\boxed{ \sf{F = \pink{ 92.16 \times }{10}^{ - 6} \: newton }}} \:$

To Find :-

→ Electrostatic force between an electron and a proton .

Explanation :-

Given that :-

• Distance between proton and electron (r)= 0.5 Å .

We know that :

$\to \red{ \boxed{ \sf{F \: = \green{ \frac{1}{4 \pi \: \epsilon_{0} } } \frac{q_{1} \times q_{2} }{ \pink{ {r}^{2} }} }} }\\ \\ \because \sf{charge \: of \: proton \: is \: equal} \\ \: \: \: \: \: \sf{ to \: charge \: of \: electron \: } \\ \\ \therefore \sf \red{ charge \: of \: both \:} = 1.6 \green{ \times {10}^{ - 19} \:} coulamb \: \\ \\ \implies \boxed{ \green{ \frac{1}{4 \pi \: \epsilon_{0} } } = 9 \times {10}^{9} \: \: N-m^{2} C^{-2} \: } \\ \\ \implies \boxed{ \sf \: 1 \: Å = {10}^{ - 10} \: m \: } \\ \sf hence \: \\ \implies \sf r \: = 0.5 \times {10}^{ - 10} \: m \: \: = 5 \times {10}^{ - 11} \: m$

Hence , required force is

$\leadsto \sf{ F = 9 \times {10}^{9} \times \frac{1.6 \times {10}^{ - 19} \times 1.6 \times {10}^{ - 19} }{25 \times {10}^{ - 22} }} \\ \\ \leadsto \sf{ \: F = 9 \times 1.6 \times 1.6}\times \frac{ {10}^{9 - 38 + 22} }{25} \\ \\ \leadsto \sf{F = \frac{9 \times 16 \times 16}{25} \times {10}^{9 - 38 + 22 - 2} } \\ \\ \leadsto \red{\boxed{ \sf{F = \pink{ 92.16 \times }{10}^{ - 6} \: newton }}}$