Question

In Fig. 21-25, four particles form a square. The charges are${\mathbf{\text{q}}}_{\mathbf{\text{1}}}\mathbf{\text{=+Q}}$,${\mathbf{\text{q}}}_{\mathbf{\text{2}}}{\mathbf{\text{=q}}}_{\mathbf{\text{3}}}\mathbf{\text{=q}}$, and${\mathbf{\text{q}}}_{\mathbf{\text{4}}}\mathbf{\text{=-2.00Q}}$. What is $\frac{q}{Q}$if the net electrostatic force on particle 1 is zero?

Short answer

The value of the net electrostatic force is$\frac{q}{Q}=\frac{1}{\sqrt{2}}=0.707$ .

Step-by-step solution

Given

The given data $q_1=+Q,q_2=q_3=q,$and$q_4=-2.00Q$

Calculate qQif the net electrostatic force on particle 1 is zero

For the net force on$q_1=+Q$to vanish, the x force component due tomust

exactly cancel the force of attraction caused by$q_4=-2.00Q$. Consequently,

$\begin{array}{c}\frac{Qq}{4\pi {\epsilon }_0{a}^2}=\frac{Q\left|2Q\right|}{4\pi {\epsilon }_0{\left(\sqrt{2}a\right)}^2}cos(45°)\\ =\frac{Q^2}{4\pi {\epsilon }_0\sqrt{2}{a}^2}\end{array}$

or$q=\frac{Q}{\sqrt{2}}$

This implies that :

$\begin{array}{c}\frac{q}{Q}=\frac{1}{\sqrt{2}}\\ =0.707\end{array}$