The electrostatic force of repulsion between two positively charged ions carrying equal charge is `3.7xx10^(-9)N`, when they are separated by a distan

Question

The electrostatic force of repulsion between two positively charged ions carrying equal charge is `3.7xx10^(-9)N`, when they are separated by a distan

The electrostatic force of repulsion between two positively charged ions carrying equal charge is `3.7xx10^(-9)N`, when they are separated by a distance of `5Å`. How much electrons are missing from each ion?
A. 10
B. 8
C. 2
D. 1

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - C
Here, `F=3.7xx10^(-9)N`
Let `q_(1)=q_(2)=q rArr r= 5Å = 5xx10^(-10)m`
`because` The force between two positively charged
`F=(1)/(4piepsilon_(0)).(q_(1)q_(2))/(r^(2))`
`rArr" "3.7xx10^(-9)=9xx10^(9)xx(qxxq)/((5xx10^(-10))^(2))`
`or q^(2)=(3.7xx10^(-9)xx25xx10^(-20))/(9xx10^(9))=10.28xx10^(-38)`
`"Charge, "q=3.2xx10^(-19)C`
Now, `q=n e rArr n=(q)/(e)=(3.2xx10^(-19))/(1.6xx10^(-19))=2`