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Three positive charges, A, B, and C, of magnitudes 3 micro C, 2 micro C, and 2 micro C...

Question:

Three positive charges, A, B, and C, of magnitudes {eq}\rm 3 \ \mu C, 2 \ \mu C, {/eq} and {eq}\rm 2 \ \mu C {/eq} respectively, are located at the corners of an equilateral triangle of side 0.2 m.

a. Find the magnitude of the force on each charge.

b. Find the magnitude of the electric field at the center of the triangle.

Coulomb's Law

We use in this answer the Coulomb's law to determine properties of a system of electrical charges.

Coulomb's law gives the force between two charges at a distance {eq}d {/eq} apart and its magnitude is given by

$$F = k \dfrac{ Q_{1} Q_{2} }{ d^{2} } \, \, , $$

where $$k = 9.0 \times 10^{9} \, \text{N} \cdot \text{m}^{2} \cdot \text{C}^{-2} \, \, . $$

The charges {eq}Q_1 {/eq} and {eq}Q_2 {/eq} may be positive or negative. The force is attractive for charges of opposite signs and repulsive for charges of equal signs.

The electric force obeys the superposition principle, which means that the resultant force on a charge due to a collection of charges is the vectorial sum of the forces due to each charge.

$$\begin{align} \mathbf{F}_{R} &= \sum_{i=1}^{N} \mathbf{F}_{i} \\ \\ &= \mathbf{F}_{1} + \mathbf{F}_{2} + \cdots + \mathbf{F}_{N} \, \, . \end{align} $$

The magnitude of the electric field produced by a charge {eq}Q {/eq} at a point {eq}P {/eq} at a distance {eq}r {/eq} away has magnitude given by

$$E = k \dfrac{ Q }{ r^{2} } \, \, , $$

and the direction of the electric field vector points away from positive charges and into negative ones.

The electric field also obeys the superposition principle. The electric field due to a collection of charges at a point of space is the vector sum of the fields produced by each of the charges,

$$\begin{align} \mathbf{E}_{R} &= \sum_{i=1}^{N} \mathbf{E}_{i} \\ \\ &= \mathbf{E}_{1} + \mathbf{E}_{2} + \cdots + \mathbf{E}_{N} \, \, . \end{align} $$

Answer and Explanation:

The charges are {eq}Q_{A} = 3 \mu\text{C} {/eq}, {eq}Q_{B} = 2 \mu\text{C} {/eq} and {eq}Q_{C} = 2 \mu\text{C} {/eq}. The distance between then is {eq}d = 0.2 \, \text{m} {/eq}. We show the setup in the diagram below.

(a) We now calculate the forces on each of the charges.

Force on the charge {eq}A {/eq}

The force on the charge {eq}A {/eq} is obtained by summing up the forces due to the other two charges, {eq}\mathbf{F}_{AB} {/eq} and {eq}\mathbf{F}_{AC} {/eq}. We sketch these forces in the diagram below.

We have

$$\begin{align*} \mathbf{F}_{A} &= \mathbf{F}_{AB} + \mathbf{F}_{AC} \\ \\ &= \left( k \dfrac{ Q_A Q_B }{ d^{2} } \right) ( - \cos 60^{\circ} \, \textbf{u}_{x} + \sin 60^{\circ} \, \textbf{u}_{y} ) + \left( k \dfrac{ Q_A Q_C }{ d^{2} } \right) ( \cos 60^{\circ} \, \textbf{u}_{x} + \sin 60^{\circ} \, \textbf{u}_{y} ) \\ \\ &= \dfrac{ k }{ d^{2} } \left[ \dfrac{ 1 }{ 2 } \left( - Q_A Q_B + Q_A Q_C \right) \, \textbf{u}_{x} + \dfrac{ \sqrt{3} }{ 2 } \left( Q_A Q_B + Q_A Q_C \right) \, \textbf{u}_{y} \right] \\ \\ &= \dfrac{ k }{ d^{2} } Q_A \left[ \dfrac{ 1 }{ 2 } \left( - \, Q_B + Q_C \right) \, \textbf{u}_{x} + \dfrac{ \sqrt{3} }{ 2 } \left( Q_B + Q_C \right) \, \textbf{u}_{y} \right] \, \, . \end{align*} $$

Applying numerical values we get

$$\begin{align} \mathbf{F}_{A} &= \dfrac{ ( 9 \times 10^{9} \, \text{N} \cdot \text{m}^{2} \cdot \text{C}^{-2} ) }{ ( 0.2 \, \text{m} )^{2} } ( 3 \mu \text{C} ) \left[ \dfrac{ 1 }{ 2 } \left( - \, 2 \mu \text{C} + 2 \mu \text{C} \right) \, \textbf{u}_{x} + \dfrac{ \sqrt{3} }{ 2 } \left( 2 \mu \text{C} + 2 \mu \text{C} \right) \, \textbf{u}_{y} \right] \\ \\ &\approx \left( 2.34 \, \text{N} \right) \, \textbf{u}_{y} \, \, . \end{align} $$

The magnitude is $$F_{A} = 2.34 \, \text{N} \, \, . $$

Force on charge {eq}\text{B} {/eq}

We sketch the forces on the charge {eq}B {/eq} due to the other charges in the diagram below

We have

$$\begin{align*} \mathbf{F}_{B} &= \mathbf{F}_{BC} + \mathbf{F}_{BA} \\ \\ &= \left( k \dfrac{ Q_B Q_C }{ d^{2} } \right) \, \textbf{u}_{x} + \left( k \dfrac{ Q_B Q_A }{ d^{2} } \right) ( \cos 60^{\circ} \, \textbf{u}_{x} - \sin 60^{\circ} \, \textbf{u}_{y} ) \\ \\ &= \dfrac{ k }{ d^{2} } Q_{B} \left[ \left( Q_C + \dfrac{1}{2} Q_A \right) \, \textbf{u}_{x} - \, \dfrac{ \sqrt{3} }{ 2 } Q_A \, \textbf{u}_{y} \right] \, \, , \end{align*} $$

and applying numerical values we get

$$\begin{align*} \mathbf{F}_{B} &= \dfrac{ ( 9 \times 10^{9} \, \text{N} \cdot \text{m}^{2} \cdot \text{C}^{-2} ) }{ ( 0.2 \, \text{m} )^{2} } ( 2 \mu \text{C} ) \left[ \left( 2 \mu \text{C} + \dfrac{3 \mu \text{C}}{2} \right) \, \textbf{u}_{x} - \, \dfrac{ \sqrt{3} \, \left( 3 \mu \text{C} \right) }{ 2 } \, \textbf{u}_{y} \right] \\ \\ &= \dfrac{ ( 9 \times 10^{9} \, \text{N} \cdot \text{m}^{2} \cdot \text{C}^{-2} ) }{ ( 0.2 \, \text{m} )^{2} } ( 2 \mu \text{C} ) \Big[ \left( 3.5 \mu \text{C} \right) \, \textbf{u}_{x} - \, \left( 2.6 \mu \text{C} \right) \, \textbf{u}_{y} \Big] \\ \\ &= ( 1.6 \, \text{N} ) \, \textbf{u}_{x} - ( 1.2 \, \text{N} ) \, \textbf{u}_{y} \, \, . \end{align*} $$

The magnitude is

$$F_{B} = \sqrt{ ( 1.6 \, \text{N} )^{2} + ( 1.2 \, \text{N} )^{2} } = 2 \, \text{N} \, \, . $$

Similarly, the force on the charge {eq}\text{C} {/eq} is the sum of the forces {eq}\mathbf{F}_{CB} {/eq} and {eq}\mathbf{F}_{CA} {/eq}, as shown in the diagram:

We have

$$\begin{align*} \mathbf{F}_{C} &= \mathbf{F}_{CB} + \mathbf{F}_{CA} \\ \\ &= - \, \left( k \dfrac{ Q_C Q_B }{ d^{2} } \right) \, \textbf{u}_{x} + \left( k \dfrac{ Q_C Q_A }{ d^{2} } \right) ( - \cos 60^{\circ} \, \textbf{u}_{x} - \sin 60^{\circ} \, \textbf{u}_{y} ) \\ \\ &= \dfrac{ k }{ d^{2} } Q_{C} \left[ - \, \left( Q_B + \dfrac{1}{2} Q_A \right) \, \textbf{u}_{x} - \, \dfrac{ \sqrt{3} }{ 2 } Q_A \, \textbf{u}_{y} \right] \, \, , \end{align*} $$

and applying values we get

$$\begin{align*} \mathbf{F}_{C} &= \dfrac{ ( 9 \times 10^{9} \, \text{N} \cdot \text{m}^{2} \cdot \text{C}^{-2} ) }{ ( 0.2 \, \text{m} )^{2} } ( 2 \mu \text{C} ) \left[ - \, \left( 2 \mu \text{C} + \dfrac{ 3 \mu \text{C} }{2} \right) \, \textbf{u}_{x} - \, \dfrac{ \sqrt{3} \, ( 3 \mu \text{C} ) }{ 2 } \, \textbf{u}_{y} \right] \\ \\ &= \dfrac{ ( 9 \times 10^{9} \, \text{N} \cdot \text{m}^{2} \cdot \text{C}^{-2} ) }{ ( 0.2 \, \text{m} )^{2} } ( 2 \mu \text{C} ) \left[ - \, \left( 3.5 \mu \text{C} \right) \, \textbf{u}_{x} - \, \left( 2.6 \mu \text{C} \right) \, \textbf{u}_{y} \right] \\ \\ &= - \, ( 1.6 \, \text{N} ) \, \textbf{u}_{x} - ( 1.2 \, \text{N} ) \, \textbf{u}_{y} \, \, . \end{align*} $$

The magnitude is

$$F_{C} = \sqrt{ ( - 1.6 \, \text{N} )^{2} + ( - 1.2 \, \text{N} )^{2} } = 2 \, \text{N} \, \, . $$

(b) The center of the triangle ( point {eq}P {/eq} ) is at a distance {eq}r = d / \sqrt{3} {/eq} of each charge.

The electric field due to the charge {eq}Q_A {/eq} is

$$\mathbf{E}_{A} = - \, k \dfrac{ Q_{A} }{ r^{2} } \, \textbf{u}_{y} \, \, , $$

the electric field due to the charge {eq}Q_B {/eq} is

$$\mathbf{E}_{B} = k \dfrac{ Q_{B} }{ r^{2} } \left( - \cos 30^{\circ} \, \textbf{u}_{x} + \sin 30^{\circ} \, \textbf{u}_{y} \right) \, \, , $$

and the electric field due to the charge {eq}Q_C {/eq} is

$$\mathbf{E}_{C} = k \dfrac{ Q_{C} }{ r^{2} } \left( \cos 30^{\circ} \, \textbf{u}_{x} + \sin 30^{\circ} \, \textbf{u}_{y} \right) \, \, . $$

The total electric field is

$$\mathbf{E} = \dfrac{ k }{ r^{2} } \left[ \dfrac{ \sqrt{3}}{ 2 } ( - \, Q_B + Q_C ) \,\textbf{u}_{x} + \left( - \, Q_A + \dfrac{Q_B}{2} + \dfrac{Q_C}{2} \right) \,\textbf{u}_{y} \right ] \, \, . $$

Since {eq}Q_B = Q_C = 2 \mu \text{C} {/eq}, the {eq}x {/eq}-component is zero. Substituting {eq}r = d / \sqrt{3} {/eq} and applying numerical values we get

$$\begin{align} \mathbf{E} &= \dfrac{ 3 \, k }{ d^{2} } \left( - \, Q_A + \dfrac{Q_B}{2} + \dfrac{Q_C}{2} \right) \,\textbf{u}_{y} \\ \\ &= \dfrac{ 3 \, ( 9 \times 10^{9} \, \text{N} \cdot \text{m}^{2} \cdot \text{C}^{-2} ) }{ ( 0.2 \, \text{m} )^{2} } \left( - \, 3 \mu \text{C} + \dfrac{ 2 \mu \text{C} }{2} + \dfrac{ 2 \mu \text{C} }{2} \right) \, \textbf{u}_{y} \\ \\ &\approx - ( 6.7 \times 10^{5} \, \text{N} / \text{C} ) \, \textbf{u}_{y} \, \, . \end{align} $$

The magnitude of the total electric field at the center of the triangle is

$$E \approx 7 \times 10^{5} \, \text{N} / \text{C} \, \, . $$


Learn more about this topic:

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Strength of an Electric Field & Coulomb's Law

from UExcel Physics: Study Guide & Test Prep

Chapter 12 / Lesson 4
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