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Found in: Page 443

### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

# (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?

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

See the step by step solution

## Step 1: 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.

## Step 2: 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, $$r = 16\;{\rm{cm}}$$

## Step 3: 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 $$2.2 \times {10^4}\;{\rm{N}}$$.