Question

Charge${\mathrm{q}}_2$in FIGURE$\mathrm{P}22.49$is in equilibrium. What is${\mathrm{q}}_1$$?$

Short answer

Charge is$-20\mathrm{nC}$.

Step-by-step solution

Calculation for force on charge

Two charged particles, one with a charge of ${\mathrm{q}}_1$.

And the other with a charge of ${\mathrm{q}}_2$, separated by a distance of$\mathrm{r}$.

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.

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

Force for charge${\mathrm{q}}_3$and${\mathrm{q}}_2$:

${\overrightarrow{\mathrm{F}}}_{32}=\frac{{\mathrm{Kq}}_3{\mathrm{q}}_2}{{\mathrm{r}}_{32}^2}$

$=\frac{\left(9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(5\times {10}^{-9}\mathrm{C}\right){\mathrm{q}}_2}{(0.1\mathrm{m}{)}^2}$

$=4500{\mathrm{q}}_2$

Calculation of charge

Force for charge ${\mathrm{q}}_1$and${\mathrm{q}}_2$

${\overrightarrow{\mathrm{F}}}_{12}=\frac{{\mathrm{Kq}}_1{\mathrm{q}}_2}{{\mathrm{r}}_{12}^2}$

$=\frac{\left(9.0\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right){\mathrm{q}}_1{\mathrm{q}}_2}{(0.2\mathrm{m}{)}^2}$

$=225\times {10}^9{\mathrm{q}}_1{\mathrm{q}}_2$

${\overrightarrow{\mathrm{F}}}_{12}=-{\overrightarrow{\mathrm{F}}}_{32}$so,

${\overrightarrow{\mathrm{F}}}_{12}=-{\overrightarrow{\mathrm{F}}}_{32}$

$225\times {10}^9{\mathrm{q}}_1{\mathrm{q}}_2=-4500{\mathrm{q}}_2$

${\mathrm{q}}_1=-\frac{4500}{225\times {10}^9}$

${\mathrm{q}}_1=-20\times {10}^{-9}\mathrm{C}$

${\mathrm{q}}_1=-20\mathrm{nC}$