Question

FIGURE $\mathrm{P}22.52$shows three charges and the net force on charge $-\mathrm{q}$. Charge$\mathrm{Q}$is some multiple $\mathrm{\alpha }$of $\mathrm{q}$. What is$\mathrm{\alpha }$$?$

Short answer

The value of the alpha is$4$.

Step-by-step solution

Calculation for horizontal force

The two particles exert forces on each other, with the force decreasing as the distance between them grows and increasing as the size of the charges increases.

Columns law of force,

${\mathrm{F}}_{\mathrm{C}}={\mathrm{F}}_{1\text{on}2}={\mathrm{F}}_{2\text{on}1}=\frac{\mathrm{K}\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$

Horizontal force between charge $-\mathrm{q}$and$-\mathrm{\alpha q}$

${\mathrm{F}}_{\mathrm{x}}=\frac{\mathrm{K}(-\mathrm{q})\left(\mathrm{\alpha q}\right)}{(2\mathrm{L}{)}^2}$

$=\frac{-{\mathrm{K\alpha q}}^2}{4{\mathrm{L}}^2}$

Calculation for alpha

Vertical force between charge$-\mathrm{q}$and$\mathrm{q}$,

${\mathrm{F}}_{\mathrm{y}}=\frac{\mathrm{K}(-\mathrm{q})\left(\mathrm{q}\right)}{{\mathrm{L}}^2}$

$=\frac{-{\mathrm{Kq}}^2}{{\mathrm{L}}^2}$

The net force forms a ${45}^{°}$angle with the horizontal axis, implying that the magnitude of the horizontal force${\mathrm{F}}_{\mathrm{x}}$is the same as the magnitude of the vertical force ${\mathrm{F}}_{\mathrm{y}}$.

So,

${\mathrm{F}}_{\mathrm{x}}={\mathrm{F}}_{\mathrm{y}}$

$\frac{-{\mathrm{K\alpha q}}^2}{4{\mathrm{L}}^2}=\frac{-{\mathrm{Kq}}^2}{{\mathrm{L}}^2}$

$\left(\frac{\mathrm{\alpha }}{4}\right)\left(\frac{-{\mathrm{Kq}}^2}{{\mathrm{L}}^2}\right)=\left(\frac{-{\mathrm{Kq}}^2}{{\mathrm{L}}^2}\right)$

$\frac{\mathrm{\alpha }}{4}=1$

$\mathrm{\alpha }=4$