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Three charges, Q_1, Q_2, and Q_3 are located in a straight line. The position of Q_2 is 0.312 m...

Question:

Three charges, {eq}Q_1, \ Q_2 {/eq}, and {eq}Q_3 {/eq} are located in a straight line. The position of {eq}Q_2 {/eq} is 0.312 m to the right of {eq}Q_1 {/eq}. {eq}Q_3 {/eq} is located 0.136 m to the right of {eq}Q_2 {/eq}.

(a) The force on {eq}Q_2 {/eq} due to its interaction with {eq}Q_3 {/eq} is directed to the (Select the corrent answer)
(i) Left, if the two charges are positive.
(ii) Left, if the two charges are negative.
(iii) Right, if the two charges are negative.
(iv) Left, if the two charges have opposite signs.
(v) Right, if the two charges have opposite signs.

(b) If {eq}Q_1 = 1.73 \ \rm \mu C {/eq}, {eq}Q_2 = -2.65 \ \rm \mu C {/eq}, and {eq}Q_3 = 3.18 \ \rm \mu C {/eq}, calculate the total force on {eq}Q_2 {/eq}.

(c) Now the charges {eq}Q_1 = 1.73 \ \rm \mu C {/eq} and {eq}Q_2 = -2.65 \ \rm \mu C {/eq} are fixed at their positions, distance 0.312 m apart, and the charge {eq}Q_3 = 3.18 \ \rm \mu C {/eq} is moved along the straight line. For what position of {eq}Q_3 {/eq} relative to {eq}Q_1 {/eq} is the net force on {eq}Q_3 {/eq} due to {eq}Q_1 {/eq} and {eq}Q_2 {/eq} zero?

Coulomb's law

The electrostatic force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance that joins them, whose magnitude is obtained by the expression:

{eq}F = \frac {1}{4 \pi \epsilon_0} \frac {|q_2 q_1|} {r^2} {/eq}

Where {eq}\frac {1} {4 \pi \epsilon_0} {/eq} it is a constant whose value is {eq}9 * 10^9 \ \frac {N m^2} {C^2} {/eq}

The magnitude of the electrostatic force is always positive however its direction depends on the signs of the charges, that is, if the charges have the same sign (negative or positive) the force is repulsive and if the charges have opposite signs the force it is of attraction.


Answer and Explanation:

(a) To choose the correct answer we have to take into account that the direction of the force depends on the signs of the charges, that is, if the charges have the same sign (negative or positive) the force is repulsive and if the charges have opposite signs the force is of attraction. So, the force on {eq}Q_2 {/eq} due to its interaction with {eq}Q_3 {/eq}is directed to the:

(i) Left, if the two charges are positive.

(ii) Left, if the two charges are negative.

(v) Right, if the two charges have opposite signs.


(b) The total force {eq}F_t {/eq} in {eq}Q_2 {/eq} results from the vector sum of the forces that individually act on each of the charges. In this case {eq}Q_1 {/eq} exerts a force {eq}F_{12} {/eq} on {eq}Q_2 {/eq} and the force exerted by {eq}Q_3 {/eq} on {eq}Q_1 {/eq} will be denoted by {eq}F_{32} {/eq} as shown in the figure:

Both values are obtained from Coulomb's Law, so replacing values we have to:

{eq}\begin {array}{cl} F_{12} &= \frac {1}{4 \pi \epsilon_0} \frac {|Q_1 Q_2|} {r^2} \\ F_{12} &= (9*10^9 \frac {Nm^2}{C^2}) \frac {|(1.73 * 10^{-6} C) (2.65 * 10^{-6} C)|}{(0.312 m)^2} \\ F_{12} &= 0.424 \ N \end {array} {/eq}


{eq}\begin {array}{cl} F_{32} &= \frac {1}{4 \pi \epsilon_0} \frac {|Q_3 Q_2|} {r^2} \\ F_{32} &= (9*10^9 \frac {Nm^2} {C^2}) \frac {|(3.18 * 10^{-6} C) (2.65 * 10^{-6} C)|} {(0.136 m)^2} \\ F_{32} &= 4.100 \ N \end {array} {/eq}


{eq}\begin {array}{cl} F_t &= 4.100 \ N - 0.424 \ N \\ F_t &= 3.676 \ N \end {array} {/eq}

The net electric force exerted on {eq}Q_2 {/eq} has a magnitude of {eq}3.676 \ N {/eq} and its direction corresponds to the positive {eq}x {/eq} axis.

(c) Charge {eq}Q_3 {/eq} could be located to the right or to the left of load {eq}Q_1 {/eq}.

We analyze the case for which it is to the right of {eq}Q_1 {/eq} where in whose position the net force on {eq}Q_3 {/eq} due to {eq}Q_1 {/eq} and {eq}Q_2 {/eq} must be zero. As we see in the figure, the senses of the forces acting on {eq}Q_3 {/eq} have the same meaning, so the total force will never be zero. Therefore it is an incorrect position.


Now we analyze the case for which it is to the left of {eq}Q_1 {/eq}. As we see in the figure the sense of the forces acting on {eq}Q_3 {/eq} have opposite meanings so that the total force can be zero for this position.


Applying Coulomb's Law to obtain both forces we have to:

The component in {eq}x {/eq} for the force exerted by {eq}Q_1 {/eq} on {eq}Q_3 {/eq} is:

{eq}\begin {array}{cl} F_{13} &= k \frac {|Q_1 Q_3|} {r^2} \\ F_{13} &= k \frac {(5.50 * 10^{-12})} {x^2} \\ \end {array} {/eq}


The component in {eq}x {/eq} for the force exerted by {eq}Q_2 {/eq} on {eq}Q_3 {/eq} is:

{eq}\begin {array}{cl} F_{23} &= - k \frac {|Q_2 Q_3|} {r^2} \\ F_{23} &= - k \frac {(8.427 * 10^{-12})} {(x+0.312)^2} \\ \end {array} {/eq}

Equating to zero the sum of both forces we have to:

{eq}\begin {array}{cl} k \frac {(5.50 * 10^{-12})} {x^2} - k \frac {(8.427 * 10^{-12})} {(x+0.312)^2} &= 0 \\ k \frac {(5.50 * 10^{-12})} {x^2} &= k \frac {(8.427 * 10^{-12})} {(x+0.312)^2} \end {array} {/eq}


We eliminate equal terms on both sides of the equation and get:

{eq}\begin {array}{cl} 8.427{x^2} &= 5.50 {(x+0.312)^2} \\ 8.427{x^2} &= 5.50 {x^2}+3.432x+0.097 \\ \end {array} {/eq}


Reducing similar terms we obtain the following quadratic equation:

{eq}\begin {array}{cl} 2.927 {x^2} - 3.432x - 0.097 &= 0 \end {array} {/eq}


Applying the quadratic formula we have:

{eq}\begin {array}{cl} x_1 &= \frac {3.432+ \sqrt {(-3.432)^2 -4(2.927)(-0.097)}}{2(2.927)} \\ x_1 &= \frac {3.432+3.593}{5.854} \\ x_1 &= 1.20 \end {array} {/eq}


{eq}\begin {array}{cl} x_2 &= \frac {3.432- \sqrt {(-3.432)^2 -4(2.927)(-0.097)}}{2(2.927)} \\ x_2 &= \frac {3.432-3.593}{5.854} \\ x_2 &= - 0.028 \end {array} {/eq}


The correct answer to locate the load {eq}Q_3 {/eq} to the left of {eq}Q_1 {/eq} is the solution for {eq}x_2 {/eq}. Therefore {eq}Q_3 {/eq} must be located at {eq}0.028 {/eq} to the left of {eq}Q_1 {/eq}.


Learn more about this topic:

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Strength of an Electric Field & Coulomb's Law

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Chapter 12 / Lesson 4
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After watching this lesson, you will be able to explain what an electric field is, distinguish between scalar and vector fields and use two electric field equations to solve problems. A short quiz will follow.


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