Question

What is the force between two small charged spheres having charges of $2 \times {10^{ - 7}}{\rm{ C}}$ and $3 \times {10^{ - 7}}{\rm{ C}}$ placed $30{\rm{ cm}}$ apart in air?

Answer 1

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. We will use this expression by rewriting it after removing the proportionality sign by constant term $\dfrac{1}{{4\pi {\varepsilon _0}}}$.

Complete step by step answer:
Given:
The value of charge on the first sphere is $2 \times {10^{ - 7}}{\rm{ C}}$.
The value of charge on the second sphere is $3 \times {10^{ - 7}}{\rm{ C}}$.
The distance between spheres is $30{\rm{ cm}} = 30{\rm{ cm}} \times \left( {\dfrac{{\rm{m}}}{{100{\rm{ cm}}}}} \right) = 0.3{\rm{ m}}$
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.
$F = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$......(1)
Here ${q_1}$ is the charge on the first sphere, ${q_2}$ is the charge on the second sphere, r is the distance between spheres and ${\varepsilon _0}$ is free space permittivity.
We know the value of free space permittivity is:
${\varepsilon _0} = 8.854 \times {10^{ - 12}}{\rm{ }}{{\rm{C}}^2}{{\rm{N}}^{ - 1}}{{\rm{m}}^{ - 2}}$
Substitute $2 \times {10^{ - 7}}{\rm{ C}}$ for ${q_1}$, $3 \times {10^{ - 7}}{\rm{ C}}$ for ${q_2}$, $0.3{\rm{ m}}$ for r and $8.854 \times {10^{ - 12}}{\rm{ }}{{\rm{C}}^2}{{\rm{N}}^{ - 1}}{{\rm{m}}^{ - 2}}$ for ${\varepsilon _0}$ in equation (1).

F = \dfrac{{\left( {2 \times {{10}^{ - 7}}{\rm{ C}}} \right)\left( {3 \times {{10}^{ - 7}}{\rm{ C}}} \right)}}{{4\pi \left( {8.854 \times {{10}^{ - 12}}{\rm{ }}{{\rm{C}}^2}{{\rm{N}}^{ - 1}}{{\rm{m}}^{ - 2}}} \right){{\left( {0.3{\rm{ m}}} \right)}^2}}}\\\ = \dfrac{{\left( {2 \times {{10}^{ - 7}}{\rm{ C}}} \right)\left( {3 \times {{10}^{ - 7}}{\rm{ C}}} \right)}}{{4 \times 3.14\left( {8.854 \times {{10}^{ - 12}}{\rm{ }}{{\rm{C}}^2}{{\rm{N}}^{ - 1}}{{\rm{m}}^{ - 2}}} \right){{\left( {0.3{\rm{ m}}} \right)}^2}}}\\\ = 5.994 \times {10^{ - 3}}{\rm{ N}} \end{array}$$ Therefore, the force between two small charged spheres having charges of $$2 \times {10^{ - 7}}{\rm{ C}}$$ and $$3 \times {10^{ - 7}}{\rm{ C}}$$ placed $$30{\rm{ cm}}$$ apart in the air is $$5.994 \times {10^{ - 3}}{\rm{ N}}$$. **Note:** We know that the force between two, unlike charge particles, is attraction force and between two like charge particles is repulsive. From the given question, we can see that both the spherical particles have positive charges; therefore, we can say that the force between them will be repulsive.