Question

What are the strength and direction of the electric field at the position indicated by the dot in FIGURE$\mathrm{EX}23.3$$?$Specify the direction as an angle above or below horizontal.

Short answer

The strength of the electric field is $7.63\times {10}^3\mathrm{V}/\mathrm{m}$.

Step-by-step solution

Angle of electric field diagram

Figure is,

Calculation for magnitude of E1

The electric field's magnitude is,

$\overrightarrow{\mathrm{E}}=\left({\mathrm{E}}_2\mathrm{cos\theta }-{\mathrm{E}}_1\mathrm{cos\theta }\right)\mathbf{i}-\left({\mathrm{E}}_2\mathrm{sin\theta }+{\mathrm{E}}_1\mathrm{sin\theta }\right)\mathbf{j}$

${\mathrm{E}}_1$:

${\mathrm{E}}_1=\frac{{\mathrm{KQ}}_1}{{\mathrm{AB}}^2}$

${\mathrm{E}}_1=\frac{\left(9\times {10}^9\mathrm{N}\right)\left(3.0\times {10}^{-9}\mathrm{C}\right)}{(5\sqrt{2}\mathrm{cm}{)}^2}$

${\mathrm{E}}_1=5.4\times {10}^3\mathrm{V}/\mathrm{m}$

Calculation for magnitude of E2

${\mathrm{E}}_2$:

${\mathrm{E}}_2=\frac{{\mathrm{KQ}}_2}{{\mathrm{BC}}^2}$

${\mathrm{E}}_2=\frac{\left(9\times {10}^9\mathrm{N}\right)\left(3.0\times {10}^{-9}\mathrm{C}\right)}{(5\sqrt{2}\mathrm{cm}{)}^2}$

${\mathrm{E}}_2=5.4\times {10}^3\mathrm{V}/\mathrm{m}$

$\mathrm{\Delta ABC},\mathrm{sin\theta }=\mathrm{cos\theta }=\frac{1}{\sqrt{2}}$

So,

localid="1649136637521" $\overrightarrow{\mathrm{E}}=\left({\mathrm{E}}_2\mathrm{cos\theta }-{\mathrm{E}}_1\mathrm{cos\theta }\right)\mathrm{i}-\left({\mathrm{E}}_2\mathrm{sin\theta }+{\mathrm{E}}_1\mathrm{sin\theta }\right)\mathrm{j}$

$\overrightarrow{\mathrm{E}}=\left(0\right)\mathrm{i}-\left(2\times \left(5.4\times {10}^3\mathrm{V}/\mathrm{m}\right)\times \frac{1}{\sqrt{2}}\right)\mathrm{j}$

$\overrightarrow{\mathrm{E}}=7.63\times {10}^3\mathrm{V}/\mathrm{m}\mathbf{j}$