Question

FIGURE $\mathrm{P}22.51$shows four charges at the corners of a square of side $\mathrm{L}$. What is the magnitude of the net force on$\mathrm{q}$?

Short answer

Magnitude of net force on charge is$(2-\sqrt{2})\frac{\mathrm{KQq}}{{\mathrm{L}}^2}$.

Step-by-step solution

Figure for magnitude of the net force on charge

Figure is,

Step 2: 

Columns law for force is,

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

$\mathrm{L}$is the distance between${\mathrm{q}}_1$andlocalid="1649103824153" $\mathrm{q}$:

${\mathrm{F}}_{{\mathrm{q}}_1\text{on}\mathrm{q}}=\frac{{\mathrm{Kq}}_1{\mathrm{q}}_{\hat{\mathrm{i}}}}{{\mathrm{L}}^2}$

$\sqrt{2}\mathrm{L}$is the distance between${\mathrm{q}}_2$andlocalid="1649103830714" $\mathrm{q}$:

${\mathrm{F}}_{{\mathrm{q}}_2\text{on}\mathrm{q}}=\frac{{\mathrm{Kq}}_{\mathrm{q}}}{(\sqrt{2}\mathrm{L}{)}^2}\cos {45}^{\mathrm{o}}\hat{\mathrm{i}}+\frac{{\mathrm{Kq}}_2\mathrm{q}}{(\sqrt{2}\mathrm{L}{)}^2}\sin {45}^{\mathrm{o}}\hat{\mathrm{j}}$

$=\left(\frac{{\mathrm{Kq}}_2\mathrm{q}}{(\sqrt{2}\mathrm{L}{)}^2}\right)\left(\frac{1}{\sqrt{2}}\right)\hat{\mathrm{i}}+\left(\frac{{\mathrm{Kq}}_2\mathrm{q}}{(\sqrt{2}\mathrm{L}{)}^2}\right)\left(\frac{1}{\sqrt{2}}\right)\hat{\mathrm{j}}$

$=\left(\frac{\mathrm{K}\left(4\mathrm{Q}\right)\mathrm{q}}{(\sqrt{2}\mathrm{L}{)}^2}\right)\left(\frac{1}{\sqrt{2}}\right)\hat{\mathrm{i}}+\left(\frac{\mathrm{K}\left(4\mathrm{Q}\right)\mathrm{q}}{(\sqrt{2}\mathrm{L}{)}^2}\right)\left(\frac{1}{\sqrt{2}}\right)\hat{\mathrm{j}}$

$=\left(\frac{4\mathrm{KQq}}{2\sqrt{2}{\mathrm{L}}^2}\right)\hat{\mathrm{i}}+\left(\frac{4\mathrm{KQq}}{2\sqrt{2}{\mathrm{L}}^2}\right)\hat{\mathrm{j}}$

Calculation for net force on charge

$\mathrm{L}$is distance for${\mathrm{q}}_3$and$\mathrm{q}$:

${\mathrm{F}}_{{\mathrm{q}}_3\text{on}\mathrm{q}}=\frac{{\mathrm{Kq}}_3\mathrm{q}}{{\mathrm{L}}^2}\hat{\mathrm{j}}$

$=\frac{-\mathrm{KQq}}{{\mathrm{L}}^2}\hat{\mathrm{j}}$

Net force is,

${\mathrm{F}}_{\text{net}}={\mathrm{F}}_{{\mathrm{q}}_1\text{on}\mathrm{q}}+{\mathrm{F}}_{{\mathrm{q}}_2\text{on}\mathrm{q}}+{\mathrm{F}}_{{\mathrm{q}}_3\text{on}\mathrm{q}}$

$=\frac{-\mathrm{KQq}}{{\mathrm{L}}^2}\hat{\mathrm{i}}+\left(\frac{4\mathrm{KQq}}{2\sqrt{2}{\mathrm{L}}^2}\right)\hat{\mathrm{i}}+\left(\frac{4\mathrm{KQq}}{2\sqrt{2}{\mathrm{L}}^2}\right)\hat{\mathrm{j}}+\frac{-\mathrm{KQq}}{{\mathrm{L}}^2}\hat{\mathrm{j}}$

$=\frac{\mathrm{KQq}}{{\mathrm{L}}^2}\left[-1+\frac{2}{\sqrt{2}}\right]\hat{\mathrm{i}}+\frac{\mathrm{KQq}}{{\mathrm{L}}^2}\left[-1+\frac{2}{\sqrt{2}}\right]\hat{\mathrm{j}}$

Magnitude of net force is,

${\mathrm{F}}_{\text{net}}=\sqrt{{\left(\frac{\mathrm{KQq}}{{\mathrm{L}}^2}\left[-1+\frac{2}{\sqrt{2}}\right]\right)}^2+{\left(\frac{\mathrm{KQq}}{{\mathrm{L}}^2}\left[-1+\frac{2}{\sqrt{2}}\right]\right)}^2}$

$=\sqrt{2}\left(\frac{\mathrm{KQq}}{{\mathrm{L}}^2}\left[-1+\frac{2}{\sqrt{2}}\right]\right)$

$=(2-\sqrt{2})\frac{\mathrm{KQq}}{{\mathrm{L}}^2}$