Question

A point charge of -3 µC is located at x = 3 m, y = -2 m. A second point charge of 12 µC is located at x = 1 m, y = 1 m.

A point charge of-3 pC is located at x = 3 m, y =-2 m. A second point charge of 12 pC İs located atx = 1 m, y 1 m (a) Find the magnitude and direction of the electric field at x1 m, y 0. × N/C X。(counterclockwise from +x axis) (b) Calculate the magnitude and direction of the force on an electron at x =-1 m, y =。 o (counterclockwise from +x axis) eBook

Answer 1

(a)

A point charge $q_1$ is located at $\vec{r}_1$. Electric field due to $q_1$ at point $\vec{r}_2$ is $\vec{E}_{21}=\frac{q_1}{4\pi\epsilon_0}\frac{(\vec{r}_2-\vec{r}_1)}{|\vec{r}_2-\vec{r}_1|^3}$

Charges and their locations are

$q_1=-3\,\mu C$ located at $\vec{r}_1=3\hat{x}-2\hat{y}$

$q_2=12\,\mu C$ located at $\vec{r}_2=1\hat{x}+1\hat{y}$

Electric field is to be calculated at $\vec{r}=-1\hat{x}+0\hat{y}$

$\vec{E}=\vec{E}_{1}+\vec{E}_{2}=\frac{q_1}{4\pi\epsilon_0}\frac{(\vec{r}-\vec{r}_1)}{|\vec{r}-\vec{r}_1|^3}+\frac{q_2}{4\pi\epsilon_0}\frac{(\vec{r}-\vec{r}_2)}{|\vec{r}-\vec{r}_2|^3}$

$\vec{E}=\frac{1}{4\pi\epsilon_0}\left[\frac{q_1(\vec{r}-\vec{r}_1)}{|\vec{r}-\vec{r}_1|^3}+\frac{q_2(\vec{r}-\vec{r}_2)}{|\vec{r}-\vec{r}_2|^3} \right ]$

$\vec{E}=\frac{1}{4\pi\epsilon_0}\left[\frac{(-3*10^{-6})(-1\hat{x}+0\hat{y}-3\hat{x}+2\hat{y})}{|-1\hat{x}+0\hat{y}-3\hat{x}+2\hat{y}|^3}+\frac{(12*10^{-6})(-1\hat{x}+0\hat{y}-1\hat{x}-1\hat{y})}{|-1\hat{x}+0\hat{y}-1\hat{x}-1\hat{y}|^3} \right ]$

$\vec{E}=\frac{1}{4\pi\epsilon_0}\left[\frac{(-3*10^{-6})(-4\hat{x}+2\hat{y})}{|-4\hat{x}+2\hat{y}|^3}+\frac{(12*10^{-6})(-2\hat{x}-1\hat{y})}{|-2\hat{x}-1\hat{y}|^3} \right ]$

$\vec{E}=\frac{1}{4\pi\epsilon_0}\left[\frac{(-3*10^{-6})(-4\hat{x}+2\hat{y})}{(2\sqrt{5})^3}+\frac{(12*10^{-6})(-2\hat{x}-1\hat{y})}{(\sqrt{5})^3} \right ]$

$\vec{E}=\frac{1}{4\pi\epsilon_0}\left[\frac{(-3*10^{-6})(-4\hat{x}+2\hat{y})}{(2\sqrt{5})^3}+\frac{(96*10^{-6})(-2\hat{x}-1\hat{y})}{(2\sqrt{5})^3} \right ]$

$\vec{E}=-\frac{1}{4\pi\epsilon_0}\left[\frac{(180\hat{x}+102\hat{y})(10^{-6})}{(2\sqrt{5})^3} \right ]$

$\vec{E}=\left[-18087.48275\,\hat{x}-10249.57356\,\hat{y} \right ]\,N/C$

Magnitude of electric field $E=|\vec{E}|=\left[(18087.48275)^2+(10249.57356)^2\right ]^{0.5}=20789.68\,N/C$

Direction of electric field makes an angle $\theta=180\degree+\tan^{-1}{\left(\frac{10249.57356}{18087.48275}\right)}=209.54\degree$ counter clock wise from

X axis.

(b)

Charge on electron $-e=-1.602*10^{-19}\,C$

Force on electron $\vec{F}=-e\vec{E}=\left[2.8976\,\hat{x}+1.642\,\hat{y} \right ]10^{-15} \,N$

Magnitude of force $|\vec{F}|=3.33*10^{-15}\,N$

Direction of force makes an angle $\theta=\tan^{-1}{\left(\frac{10249.57356}{18087.48275}\right)}=29.54\degree$

counter clock wise from X axis.