Question

Part A In the figure, charge q 1-2.8 × 10-6 C is placed at the origin and charge q 2 = 5.0 x 106 C is placed on the x-axis at x = a third charge Q--8.3 C be placed such that the resultant force on this third charge is zero? 20 m. Where a ong the positive x-axis can the resultantforcemme qrdor e 0.20 m 92 41 Express your answer using two significant figures

Answer 1

Let the third charge be placed at a distance x to the right of the origin on the positive side of the x-axis.

Then the force F₁ due to charge q₁on charge Q=$q_{1}Q/4\pi \epsilon _{0}x^{2}=2.8*10^{-6}*8.3*10^{-6}*9*10^{9}/x^{2}=209*10^{-3}/x^{2}$where $1/4\pi \epsilon _{0}=9*10^{9}Nm^{2}/C^{2}$

Similarly the force F₂ due to charge q₂ on charge Q=$q_{2}Q/4\pi \epsilon _{0}(x+0.2)^{2}=5*10^{-6}*8.3*10^{-6}*9*10^{9}/(x+0.2)^{2}=374*10^{-3}/(x+0.2)^{2}$./

This force F₂ points to the right while the force F₁ points to the left.

Since the resultant force F=0, F₁=F₂

_{i.e. $374*10^{-3}/(x+0.2)^{2}=209*10^{-3}/x^{2}$ i.e.}

_{$((x+0.2)/x)^{2}=374/209=1.8$ i.e $(x+0.2)/x=\sqrt{1.8}=1.3$ .}

_{i.e $1.3x-x=0.2$ or $x=0.2 /0.3=0.67 m.$}

Thus the third charge Q must be placed at a distance x=0.67m to the right of the origin for the resultant force on it due to the other 2 charges to be zero.