Answer to Question #174419 in Electric Circuits for JC young

Question

find the net electric force on charge 1, due to the other charges.

q1= +3.0x10^-5

q2= -5.0x10^-5

q3= -5.0x10^-5

distance between q1 and q2 is 30cm

distance between q1 and q3 is 30cm

Q3 is northern most point, q2 is the eastern most point

q2 is the western most point.

Answer 1

Explanations & Calculations

• Let's consider the charge q3 90 degrees to the north of q1: at the top (north) & q2 is 90 degrees to the right of q1: (east).
• Then the goal is to calculate the force on q1 from the other two: q2 & q3.

• From Coulomb's law, force on q1 from q2,

$\qquad\qquad \begin{aligned} \small F_1&= \small k\frac{|q_1||q_2|}{r^2}\\ &=\small 9\times10^{-9}Nm^2C^{-2}\cdot\frac{3\times10^{-5}\times5\times10^{-5}}{0.3^2m^2}\\ &=\small 1.5\times10^{-16}N\,[\text{attractive: to the east towards q2}] \end{aligned}$

• Force on q1 fro q3

$\qquad\qquad \begin{aligned} \small F_2&=\small 9\times10^{-9}\cdot\frac{(3\times10^{-5})\times(5\times10^{-5})}{0.3^2}\\ &=\small 1.5\times10^{-16}N\,[\text{attractive: to the North towards q3}] \end{aligned}$

• As it can be seen $\small |F_1|=|F_2|$ and they are orthogonal.
• Then it is about finding the resultant of those two to calculate the net electric force on q1.
• Then,

$\qquad\qquad \begin{aligned} \small F_{net}&=\small \sqrt{|F_1|^2+|F_2|^2}\\ &=\small \sqrt{2|F_1|^2}\\ &=\small \sqrt2\cdot|F_1|\\ &=\small\sqrt2\times1.5\times10^{-16}\,N\\ &=\small \bold{2.12\times10^{-16}\,N}\\ \\ \small \theta&=\small\tan^{-1}\frac{|F_1|}{|F_2|}=\tan^{-1}(1)=45^0 \end{aligned}$

• Therefore, the net force is,

$\qquad\qquad\small F_{net}=2.12\times10^{-16}N[45^0 N \,of\,E]$