What's the force on charge Q3?

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What's the force on charge Q3?

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Answer 1 · 2017-11-05T03:02:07

$394 N$ $394"N"$ to the left.

Explanation:

The force between two point charges is given by Coulomb's Law:

$F = \frac{1}{4 π ε_{0}} \cdot \frac{| q_{1} | | q_{2} |}{r^{2}}$ $color(blue)(F=1/(4piepsilon_0)*(abs(q_1)abs(q_2))/r^2)$

where $r$ $r$ is the distance between the charges

To find the net force on $Q_{3}$ $Q_3$ , we will need to calculate the force exerted on it by both $Q_{1}$ $Q_1$ and $Q_{2}$ $Q_2$ .

We have the following information:

$Q_{1} = - 50 μ C$ $Q_1=-50muC$
$Q_{2} = 100 μ C$ $Q_2=100muC$
$Q_{3} = - 20 μ C$ $Q_3=-20muC$
$r_{12} = 20.0 cm$ $r_(12)=20.0"cm"$
$r_{13} = 40.0 cm$ $r_(13)=40.0"cm"$
$ε_{0} = 8.85 10^{- 12} C^{2} / {Nm}^{2}$ $epsilon_0=8.85 10^(−12)" C"^2//"Nm"^2$

Let's start with $Q_{1}$ $Q_1$ .

$Q_{1}$ $Q_1$ and $Q_{3}$ $Q_3$ are both negative charges. Therefore the charges repel, and the force exerted on $Q_{3}$ $Q_3$ by $Q_{1}$ $Q_1$ points to the right.

$F_{13} = \frac{1}{4 π ε_{0}} \cdot \frac{∣ ∣ - 50 \cdot 10^{- 6} C ∣ ∣ ∣ ∣ - 20 \cdot 10^{- 6} C ∣ ∣}{{(40.0 \cdot 10^{- 2} m)}^{2}}$ $F_(13)=1/(4piepsilon_0)*(abs(-50*10^-6C)abs(-20*10^-6C))/(40.0*10^-2m)^2$

$\Rightarrow \approx 56 N$ $=>~~56"N"$ (to the right)

**Now we can calculate the force between $Q_{2}$ $Q_2$ and $Q_{3}$ $Q_3$ : **

$Q_{2}$ $Q_2$ is a positive charge, whereas $Q_{3}$ $Q_3$ is a negative charge. Therefore the charges attract, and the force exerted on $Q_{3}$ $Q_3$ by $Q_{2}$ $Q_2$ is to the left.

$F_{23} = \frac{1}{4 π ε_{0}} \cdot \frac{∣ ∣ 100 \cdot 10^{- 6} C ∣ ∣ ∣ ∣ - 20 \cdot 10^{- 6} C ∣ ∣}{{(20.0 \cdot 10^{- 2} m)}^{2}}$ $F_(23)=1/(4piepsilon_0)*(abs(100*10^-6C)abs(-20*10^-6C))/(20.0*10^-2m)^2$

$\Rightarrow \approx 450 N$ $=>~~450"N"$ (to the left)

Then, since the charges are all placed along a straight line, we can simply add these values for force together to get the net force.
However, note that these are forces and do have directions associated with them. We can already see that the net force will be to the left, as this is is a much greater value than that to the right.
We will therefore subtract $F_{13}$ $F_(13)$ from $F_{23}$ $F_(23)$ .

$F_{on Q_{3}} = 450 N - 56 N=394 N$ $F_(" on "Q_3)=450"N"-56"N=394"N"$

That is, $394 N$ $394"N"$ to the left.

Alternatively, you can define left as the negative direction and make the calculated force net negative. Then upon adding the two forces you would obtain a negative net force, which indicates that the net force is to the left.

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What's the force on charge Q3?

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