Question

(I) What is the magnitude of the force a\({\bf{ + 25}}\;{\bf{\mu C}}\)charge exerts on a\({\bf{ + 2}}{\bf{.5}}\;{\bf{mC}}\)charge 16 cm away?

Short answer

The magnitude of the force exerted is \(2.2 \times {10^4}\;{\rm{N}}\).

Step-by-step solution

Understanding the electric force

A point charge exerts an electric force on another point charge in the vicinity. It could be attractive as well as repulsive depending upon the nature of the two charges.

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 first charge is, \({Q_1} = + 25\;{\rm{\mu C}}\)

The second charge is \({Q_2} = 2.5\;{\rm{mC}}\)

The separation between the charges is, \(r = 16\;{\rm{cm}}\)

Determination of magnitude of the force

The expression for the electric force is given as:

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

Substitute the values in the above expression.

\(\begin{aligned}{l}F = \left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right)\frac{{\left( {25 \times {{10}^{ - 6}}\;{\rm{C}}} \right)\left( {2.5 \times {{10}^{ - 3}}\;{\rm{C}}} \right)}}{{{{\left( {16\;{\rm{cm}} \times \frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}} \right)}^2}}}\\F = 2.2 \times {10^4}\;{\rm{N}}\end{aligned}\)

Thus, the magnitude of the force exerted is\(2.2 \times {10^4}\;{\rm{N}}\).