Question

The net force on the$1.0\mathrm{nC}$charge in FIGURE$\mathrm{P}22.48$is zero What is$\mathrm{q}$$?$

Short answer

Charge with net force is$0.68\mathrm{nC}$.

Step-by-step solution

Figure for charge on net force

Figure is ,

Calculation for charge

Horizontal components are cancelled out by each other.

The vertical component's force is upward.

Because the web force on$\mathrm{q}$is zero.

$\mathrm{q}$must be negative charged, causing to fall.

$2\mathrm{Fcos\theta }={\mathrm{F}}_{\mathrm{e}}$

$\left(\mathrm{cos\theta }=\frac{2}{\sqrt{13}}\right)$

$2\times \frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\times \frac{\left(2\right)\left(1\right)}{13}\times \frac{2}{\sqrt{13}}=\frac{1}{4{\mathrm{\pi ϵ}}_{\mathrm{o}}}\times \frac{\left(1\right)\left(\mathrm{q}\right)}{(2{)}^2}$

$\mathrm{q}=\frac{8\times 4}{13\sqrt{3}}=0.68\mathrm{nc}$