Question

A positive point charges$\mathrm{Q}$located at$\mathrm{x}=\mathrm{a}$and a negative point charge $-\mathrm{Q}$is at$\mathrm{x}=-\mathrm{a}$.A positive charge$\mathrm{q}$can be place anywhere on the$\mathrm{y}$- axis. Find an expression for${\left({\mathrm{F}}_{\text{net}}\right)}_{\mathrm{x}}$, the component of the net force on$\mathrm{q}$.

Short answer

The component of the net force on charge is$\left(\frac{-2\mathrm{KqQa}}{{\left({\mathrm{y}}^2+{\mathrm{a}}^2\right)}^{(3/2)}}\right)$.

Step-by-step solution

Figure for net force on charge

Figure is,

Calculation for 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.

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

The electrostatic constant is localid="1649101048148" $9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2$.

The charge $\mathrm{q}$is a long way away.

$\mathrm{r}=\sqrt{{\mathrm{y}}^2+{\mathrm{a}}^2}$

$-\mathrm{Q}$and$+\mathrm{Q}$from each charge It's on the $\mathrm{y}$- axis as well.

As a result, the force between $\mathrm{q}$and $\mathrm{Q}$is equal to

${\mathrm{F}}_1=\mathrm{Fcos\theta }=\left(\frac{-\mathrm{KqQ}}{{\mathrm{r}}^2}\right)\mathrm{cos\theta }$

The force between$\mathrm{q}$and $\mathrm{Q}$can be calculated as follows:

${\mathrm{F}}_2=-\mathrm{Fcos\theta }=-\left(\frac{\mathrm{KqQ}}{{\mathrm{r}}^2}\right)\mathrm{cos\theta }$

Calculation for net force

The term

$\mathrm{cos\theta }=\frac{\mathrm{a}}{\mathrm{r}}$

Two forces ${\mathrm{F}}_1$and ${\mathrm{F}}_2$:

${\mathrm{F}}_2={\mathrm{F}}_1=\left(\frac{-\mathrm{KqQ}}{{\mathrm{r}}^2}\right)\mathrm{cos\theta }$

$=\left(\frac{-\mathrm{KqQ}}{{\mathrm{r}}^2}\right)\left(\frac{\mathrm{a}}{\mathrm{r}}\right)$

$=\left(\frac{-\mathrm{KqQa}}{{\mathrm{r}}^3}\right)$

$=\left(\frac{-\mathrm{KqQa}}{{\left(\sqrt{{\mathrm{y}}^2+{\mathrm{a}}^2}\right)}^3}\right)$

$=\left(\frac{-\mathrm{KqQa}}{{\left({\mathrm{y}}^2+{\mathrm{a}}^2\right)}^{(3/2)}}\right)$

Net force is,

${\mathrm{F}}_{\text{net}}={\mathrm{F}}_1+{\mathrm{F}}_2$

$=\left(\frac{-\mathrm{KqQa}}{{\left({\mathrm{y}}^2+{\mathrm{a}}^2\right)}^{(3/2)}}\right)+\left(\frac{-\mathrm{KqQa}}{{\left({\mathrm{y}}^2+{\mathrm{a}}^2\right)}^{(3/2)}}\right)$

$=\left(\frac{-2\mathrm{KqQa}}{{\left({\mathrm{y}}^2+{\mathrm{a}}^2\right)}^{(3/2)}}\right)$