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Expert-verified Found in: Page 716 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In Fig. 24-72, two particles of charges q1 and q2 are fixed to an x-axis. If a third particle, of charge ${\mathbf{+}}{\mathbf{6}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\mu C}}$, is brought from an infinite distance to point P, the three-particle system has the same electric potential energy as the original two-particle system. What is the charge ratio ${{\mathbf{q}}}_{{\mathbf{1}}}{\mathbf{/}}{{\mathbf{q}}}_{{\mathbf{2}}}$? The charge ratio ${\mathrm{q}}_{1}/{\mathrm{q}}_{2}$ is -1.66.

See the step by step solution

## Step 1: Given data:

Draw the given figure as below. Here, q1 and q2 are fixed charges.

The charge, ${\mathrm{q}}_{3}=+6.0\mathrm{\mu C}=+6.0×{10}^{-6}\mathrm{C}$

## Step 2: Determining the concept:

Use the electric potential energy for the final three particle system,

$\begin{array}{rcl}{{\mathbf{U}}}_{{\mathbf{net}}}& {\mathbf{=}}& {{\mathbf{U}}}_{{\mathbf{1}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{2}}}{\mathbf{+}}{{\mathbf{U}}}_{{\mathbf{3}}}\\ & {\mathbf{=}}& \frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\left[\frac{{q}_{1}{q}_{2}}{d}+\frac{{q}_{2}{q}_{3}}{\left(1.5d\right)}+\frac{{q}_{1}{q}_{3}}{\left(2.5d\right)}\right]\end{array}$

If two like charges (two protons or two electrons) are brought towards each other, the potential energy of the system increases. If two unlike charges i.e. a proton and an electron are brought towards each other, the electric potential energy of the system decreases.

Formulae:

The electric potential is define by,

${\mathrm{U}}_{12}=\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}.\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{\mathrm{d}}$

The bet potential is define by,

$\begin{array}{rcl}{\mathrm{U}}_{\mathrm{net}}& =& {\mathrm{U}}_{1}+{\mathrm{U}}_{2}+{\mathrm{U}}_{3}\end{array}$

Where, U is electric potential energy, R is distance between the point charges, q is charge, ${\mathrm{\epsilon }}_{0}$ is the permittivity of free space, and k is the Coulomb’s constant having a value

## Step 3: Determining the charge ratio of

The net potential energy is,

$\begin{array}{rcl}{\mathrm{U}}_{\mathrm{net}}& =& {\mathrm{U}}_{1}+{\mathrm{U}}_{2}+{\mathrm{U}}_{3}\\ & =& \frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\left[\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{\mathrm{d}}+\frac{{\mathrm{q}}_{2}{\mathrm{q}}_{3}}{\left(1.5\mathrm{d}\right)}+\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{3}}{\left(2.5\mathrm{d}\right)}\right]\end{array}$

According to the given condition,

$\begin{array}{rcl}\frac{1}{4{\mathrm{\pi \epsilon }}_{0}}\left[\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{2}}{\mathrm{d}}+\frac{{\mathrm{q}}_{2}{\mathrm{q}}_{3}}{\left(1.5\mathrm{d}\right)}+\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{3}}{\left(2.5\mathrm{d}\right)}\right]& =& \frac{1}{4{\mathrm{\pi \epsilon }}_{0}}.\frac{1}{\mathrm{d}}\left[{\mathrm{q}}_{1}{\mathrm{q}}_{2}\right]\\ {\mathrm{q}}_{1}{\mathrm{q}}_{2}+\frac{{\mathrm{q}}_{2}{\mathrm{q}}_{3}}{1.5\mathrm{d}}+\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{3}}{2.5\mathrm{d}}& =& {\mathrm{q}}_{1}{\mathrm{q}}_{2}\\ \frac{{\mathrm{q}}_{2}{\mathrm{q}}_{3}}{1.5\mathrm{d}}& =& -\frac{{\mathrm{q}}_{1}{\mathrm{q}}_{3}}{2.5\mathrm{d}}\\ \frac{{\mathrm{q}}_{1}}{{\mathrm{q}}_{2}}& =& -1.66\end{array}$

Hence, the charge ratio ${\mathrm{q}}_{1}/{\mathrm{q}}_{2}$ is -1.66. ### Want to see more solutions like these? 