Question

The electrostatic force of repulsion between two positively charged ions carrying equal charges is 3.7*10^-9 N , when they are separated by distance of 5 Amg. How many electrons are missing from each ions?

Thanks! _/\_​

Answer 1

Given,

Force between the two charges = $3.7*10^{-9}N$

Distance between the charges, r = 5 А° = $5*10^{-10}m$

q1 = q2 =q

We know,

F = $\frac{1}{4\pi\epsilon_0}\frac{q^{2} }{r^{2} }$

on putting the values,

$3.7*10^{-9} = 9*10^{9} *\frac{q^{2}}{(5*10^{-10})^{2}}$

$\frac{3.7*10^{-9}*25*10^{20}}{9*10^{9}} = q^{2}$

On solving we get,

$q^{2} = 10.28*10^{-38} \\\\q = 3.20*10^{-19} C$

Thus, Charge = $3.20*10^{-19} C$

Now we know,

Q = $\pm$ne              [here charge is positive]

on putting value,

$3.20*10^{-19} = n *1.6*10^{-19}\\\\n = \frac{3.20}{1.6}$

n = 2

Thus , the number of missing electrons is 2.

Answer 2

Explanation:

Let us assume the charges be Q

F = K Q^2/r^2

k = 9 * 10^9 Nm^2C^-2

F = 3.7 * 10^-9 N

Radius = 5 * 10 ^-10 m

[ 1 Amg = 1 * 10^-10 m]

after putting all values in formula we get,

$3.7 \times 10 {}^{ - 9} = \frac{9 \times 10 {}^{9 \times q {}^{2} } }{25 \times 10 {}^{ - 20} } \\ \\ \\ \\ q = 3.19 \times 10 {}^{ - 19} \\ q = - ne \\ n = \frac{ - q}{e} \\ = \frac{ - 3.19 \times 10 {}^{ - 19} }{ - 1.6 \times 10 {}^{ } } \\ \\ = 1.9$

Approx

n = 2