Question

(II) Two charged dust particles exert a force of\({\bf{4}}{\bf{.2 \times 1}}{{\bf{0}}^{{\bf{ - 2}}}}\;{\bf{N}}\)on each other. What will be the force if they are moved so they are only one-eighth as far apart?

Short answer

The force exerted when dust particles moved to one-eighth as far apart is \(2.7\;{\rm{N}}\).

Step-by-step solution

Understanding the relation between electric force and separation distance

The electric force exerted between two charges is directly proportional to the magnitude of charges and inversely proportional to the square of the separation distance.

The expression for the electric force is given as:

\(F = k\frac{{{Q_1}{Q_2}}}{{{r^2}}}\) … (i)

Here, k is the Coulomb’s constant,\({Q_1},\;{Q_2}\)are the charges and r is the separation between the charges.

Given Data

The force exerted on dust particles is, \(F = 4.2 \times {10^{ - 2}}\;{\rm{N}}\).

Determination of force if the charged particles move 1/8 distance

From equation (i), the electric force is,

\(F \propto \frac{1}{{{r^2}}}\)

When the distance is multiplied by the factor \(\left( {\frac{1}{8}} \right)\), due to the square of the distance, the force is multiplied by 64.

Therefore, the new force exerted between the dust particles is,

\(\begin{aligned}{l}F' = 64F\\F' = 64\left( {4.2 \times {{10}^{ - 2}}\;{\rm{N}}} \right)\\F' \approx 2.7\;{\rm{N}}\end{aligned}\)

Thus, the force exerted when dust particles moved to one-eighth as far apart is\(2.7\;{\rm{N}}\).