Answer to Question #89882 in Electric Circuits for michael

Question

Three charges lie along the x-axis. the positive charge q1=15µC is at x=2.0m, and the positive charge q2=6.0µC is at origin. Where must a negative charge q3 must be placed on the x axis so that the resultant electric force on it is zero?

Answer 1

According to Coulomb’s law the force F between two point charges, q₁ and q₂, separated by a distance r is equal to

$$F=k\frac{\left| {{q}_{1}}{{q}_{2}} \right|}{{{r}^{2}}}$$

q₁=15µC, q₂=6.0µC, r=2m. Let r₁₃ and r₂₃ are the distances between q₁ and q₃ as well as between q₂ and q_3, respectively, r₁₃ + r₂₃=r=2.

For net force to be zero on q₃ must be performed

$${{\vec{F}}_{net}}={{\vec{F}}_{13}}+{{\vec{F}}_{23}}=0$$

or

$${{\vec{F}}_{13}}=-{{\vec{F}}_{23}}$$

For absolute values of forces we get

$$k\frac{{{q}_{1}}{{q}_{3}}}{r_{13}^{2}}=k\frac{{{q}_{2}}{{q}_{3}}}{r_{23}^{2}}$$

or

$$\frac{{{q}_{1}}}{{{\left( r-{{r}_{23}} \right)}^{2}}}=\frac{{{q}_{2}}}{r_{23}^{2}}$$

Solve this equation for   r₂₃

$$\frac{{{\left( r-{{r}_{23}} \right)}^{2}}}{r_{23}^{2}}=\frac{{{q}_{1}}}{{{q}_{2}}}\,\,\,\Rightarrow \,\,\,\frac{r-{{r}_{23}}}{{{r}_{23}}}=\sqrt{\frac{{{q}_{1}}}{{{q}_{2}}}}$$

Then

$${{r}_{23}}=\frac{r}{1+\sqrt{\frac{{{q}_{1}}}{{{q}_{2}}}}}$$

Substitute known values

$${{r}_{23}}=\frac{2\,m}{1+\sqrt{\frac{15\mu C}{6.0\mu C}}}=0.77\,meters$$

Since the charge q₂ is at the origin, then the coordinate of charge q₃ is x₃=0.77meters