Question

ten electrons have been removed from each atom to form ions. find electrostatic force between two such ions when separated by distance of four angstrom in dielectric constant 4

Answer 1

The force is $3.6 \times 10^{-8} \mathrm{N}.$

Explanation:

As per coulomb’s law the ‘electrostatic force’ between two charges is given by the equation,

$\mathrm{F}_{\mathrm{e}}=\frac{1}{4 \pi \varepsilon} \cdot \frac{Q 1 Q 2}{r^{2}}$

Here both given charges are equal i.e. $10 \times 1.6 \times 10^{-19}=16 \times 10^{-19}$

The “r” here is the distance between two charges and ε is ‘permittivity’ of the medium which is equal to k times vacuum’s permittivity. Now $\frac{1}{4 \pi \varepsilon}=9.10^{9}$ is coulomb’s constant.

Putting the values and solving the equation,

we get the answer as $3.6 \times 10^{-8} \mathrm{N}$.

Answer 2

Answer:

answer is in the photo...

hope it helps