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Two point charges are placed on the x-axis as follows: Charge q1=+4.00nCq _ { 1 } = + 4.00 n C is located at x = 0.200 m, and charge q2=+5.00nCq _ { 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 q3=6.00nC\mathrm { q } _ { 3 } = 6.00 nC that is placed at the origin?

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To find the force on q3q_{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 Q1Q_{1} and Q2Q_{2} on distance rr:

FCoulomb=14πϵ0Q1Q2r2F_{\text{Coulomb}}= \frac{1}{4\pi\epsilon_{0}}\frac{|Q_{1}Q_{2}|}{r^{2}}

ϵ0=8.8541012C2Nm2\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:

14πϵ0=9.0109Nm2C2\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=4.0μCQ=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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