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# Two point charges are placed on the x-axis as follows: Charge $q _ { 1 } = + 4.00 n C$ is located at x = 0.200 m, and charge $q _ { 2 } = + 5.00 n C$ is at x = -0.300m. What are the magnitude and direction of the total force exerted by these two charges on a negative point charge $\mathrm { q } _ { 3 } = 6.00 nC$ that is placed at the origin?

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To find the force on $q_{3}$, we need to calculate the magnitudes and direction of the force for each interaction, and then sum them accordingly.

We will calculate all the forces according to Coulomb's law, which describes the force between two charges $Q_{1}$ and $Q_{2}$ on distance $r$:

$F_{\text{Coulomb}}= \frac{1}{4\pi\epsilon_{0}}\frac{|Q_{1}Q_{2}|}{r^{2}}$

$\epsilon_{0}=8.854\cdot 10^{-12} \,\mathrm{\frac{C^{2}}{N\cdot m^{2}}}$ is the electric permitivity of vacuum. We will use an approximate value of the whole constant factor:

$\frac{1}{4\pi\epsilon_{0}}=9.0\cdot 10^{9}\,\mathrm{\frac{N\cdot m^{2}}{C^{2}}}$

The net force on charge $Q=4.0\,\mathrm{\mu C}$ we will get using the concept of superposition, which allows us to calculate the net force by simply summing the force of each charge interacting with charge Q.

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