Answer to Question #287818 in Electric Circuits for Faaaa

Question

Two charges q1 = -6 µC and q2 = +10 µC are placed 12 cm apart. Another charge, q=-2 μC is placed in the middle. Determine the magnitude and direction of the force acting on q3 by the charge q1 and q2.

Answer 1

Explanations & Calculations

• It is needed to use Coulomb's law to the 2 charge combinations described here: $\small (-6\mu C\,,\, -2\mu C)\,\&\,(+10\mu C\, , \, -2\mu C)$

$\qquad\qquad \begin{aligned} \small F_1&=\small k\frac{q_1.q_2}{r^2}\\ &=\small (9\times10^9\,Nm^2C^{-2}).\frac{(-6\times10^{-6} C)\times (-2\times 10^{-6}C)}{(0.06\,m)^2}\\ &=\small +30\,N\\ \\ \small F_2&=\small (9\times10^9).\frac{(+10\times10^{-6})\times(-2\times10^{-6})}{(0.06)^2}\\ &=\small -50\,N \end{aligned}$

• The signs of the forces are different, meaning that the forces are acting not in the same direction rather opposite to each other.
• Then the total is $\small F_{net}=30-50=-20\,N$.
• If we consider the arrangement is in a way similar to $\small -6\mu C \to -2\mu C\to+10\mu C$, then the positive side is the left.
• Therefore, the net force on the q3 is to the right $\small -2\mu C \to 10\mu C$ .