Question

If a= 3.0mm, b= 4.0mm, Q1= 60 nC, Q2= -80 nC and q= 80 nC in the figure below, what is the magnitude of the total electric force on q?

Answer 1

The figure shows the arrangement of charges and forces acting on them.

Given that the charges are \(\mathrm{Q}_{1}=60 \mathrm{n} \mathrm{C}, \mathrm{Q}_{2}=-80 \mathrm{n} \mathrm{C}\) and \(\mathrm{q}=80 \mathrm{nC}\)

The electric force on q due to \(\mathrm{Q}_{1}\)

$$ \begin{aligned} F_{1} &=k \frac{Q_{1} q}{r^{2}} \\ &=\left(9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{\left(60 \times 10^{-9} \mathrm{C}\right)\left(80 \times 10^{-9} \mathrm{C}\right)}{\left(7 \times 10^{-3} \mathrm{~m}\right)^{2}} \\ &=0.882 \mathrm{~N}(i) \end{aligned} $$

The electric force on \(q\) due to \(Q_{2}\)

$$ \begin{aligned} F_{2} &=k \frac{Q_{1} q}{r^{2}} \\ &=\left(9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{\left(80 \times 10^{-9} \mathrm{C}\right)\left(80 \times 10^{-9} \mathrm{C}\right)}{\left(4 \times 10^{-3} \mathrm{~m}\right)^{2}} \\ &=3.6 \mathrm{~N}(-i) \end{aligned} $$

The net force is

$$ =3.6 \mathrm{~N}(-i)+0.882 \mathrm{~N}(i) $$

$$ =-2.7 \mathrm{~N}(i) $$

Therefore the magnitude of force is \(2.7 \mathrm{~N}\)