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Expert-verified Found in: Page 443 ### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922 # (III) Two charges, $$- {\bf{Q}}$$ and $$- {\bf{3Q}}$$ are a distance l apart. These two charges are free to move but do not because there is a third (fixed) charge nearby. What must be the magnitude of the third charge and its placement in order for the first two to be in equilibrium?

The magnitude of the third charge is $$0.40Q$$ at a distance $$0.37l$$ from the $$- Q$$ charge in between the two charges.

See the step by step solution

## Understanding the force between two point charges

The force between two point charges relies on the magnitude of both the charges and the separation between them.

The expression for the force between two point charges 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 them.

In an equilibrium state, the net force on each charge must be zero.

## Given data

The given two charges are $$- Q$$ and $$- 3Q$$.

The distance between the charges is $$l$$.

## Determination of the position of the third charge

Let there is q charge at a distance of x from the charge $$- Q$$ in between the two charges. The whole system is in equilibrium, then the force on each charge is zero.

For the equilibrium of the $$- Q$$ charge, you can get,

\begin{aligned}{c}k\frac{{qQ}}{{{x^2}}} = k\frac{{3Q \times Q}}{{{l^2}}}\\\frac{q}{{{x^2}}} = \frac{{3Q}}{{{l^2}}}\\q = 3Q\frac{{{x^2}}}{{{l^2}}}\end{aligned} … (i)

Now, for the equilibrium of the charge q you get,

\begin{aligned}{c}k\frac{{qQ}}{{{x^2}}} = k\frac{{q \times 3Q}}{{{{\left( {l - x} \right)}^2}}}\\3{x^2} = {\left( {l - x} \right)^2}\\x = \frac{l}{{\sqrt 3 + 1}}\\x = 0.37l\end{aligned}

## Determination of the magnitude of the third charge

From equation (i), the magnitude of the third charge is,

\begin{aligned}{c}q = 3Q\frac{{{{\left( {\frac{l}{{\sqrt 3 + 1}}} \right)}^2}}}{{{l^2}}}\\ = 3Q \times \frac{{{l^2}}}{{{{\left( {\sqrt 3 + 1} \right)}^2}}} \times \frac{1}{{{l^2}}}\\ = \frac{3}{{{{\left( {\sqrt 3 + 1} \right)}^2}}}Q\\ = 0.40Q\end{aligned}

Thus, the magnitude of the third charge is $$0.40Q$$ at a distance $$0.37l$$ from the $$- Q$$ charge in between the two charges. ### Want to see more solutions like these? 