Question

What will be the force of electric repulsion between two small spheres placed 1 m apart, if each has a deficit of 10^8 electrons ?​

Answer 1

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Answer 2

Answer:

Force between the two small charged spheres is $6 \times 10^{-3} \mathrm{~N}$. The charges are of same nature.

Explanation:

The force between two small charged spheres is directly proportional to the charge on both the spheres and inversely proportional to the square of the distance between them.

Given,

Charge on the first sphere, $\mathrm{q}_1=2 \times 10^{-8} \mathrm{C}$

Charge on the second sphere, $\mathrm{q}_2=3 \times 10^{-8} \mathrm{C}$

Distance between the spheres, $30 \mathrm{~cm}=30 \mathrm{~cm} \times\left(\frac{\mathrm{m}}{100 \mathrm{~cm}}\right)=0.3 \mathrm{~m}$

We are required to calculate the value of force between the given two spheres due to their charge.

Let us write the expression for force of attraction or repulsion between the given spheres.

$\mathrm{F}=\frac{\mathrm{q}_1 \mathrm{q}_2}{4 \pi \varepsilon_0 \mathrm{r}^2}$

Where, $\varepsilon_0=$ Permittivity of free space

$& \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2}$

$& \mathrm{~F}=\frac{9 \times 10^9 \times 2 \times 10^{-8} \times 3 \times 10^{-8}}{(0.3)^2}=6 \times 10^{-3} \mathrm{~N}$

Hence, force between them will be repulsive.

Hence, force between the two small charged spheres is $6 \times 10^{-3} \mathrm{~N}$. The charges are of same nature.

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