Question

1. A charge q241 uc is located at (8,0) and q3- -41 uc is a. Calculate the electric field at the origin (0,0). (magnitude b. Calculate the force on a -2.10 uc charge if it is placed at located at (O,12) and angle) the origin. (magnitude and angle)

Answer 1

Electric field at $\vec{r}$ at  due to a charge $q_1$ positioned at  $\vec{r}_1$  is $\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q_1(\vec{r}-\vec{r}_1)}{|\vec{r}-\vec{r}_1|^3}$

241 C
located at  $\vec{r}_2=8\hat{i}\,m$

$q_3=-41\,\mu C$ located at  $\vec{r}_3=12\hat{j}\,m$

$\vec{r}=0$ (origin)

a) Electric field at origin is  $\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q_2(\vec{r}-\vec{r}_2)}{|\vec{r}-\vec{r}_2|^3}+\frac{1}{4\pi\epsilon_0}\frac{q_3(\vec{r}-\vec{r}_3)}{|\vec{r}-\vec{r}_3|^3}$

$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{(-41*10^{-6})(0-8\hat{i})}{|0-8\hat{i}|^3}+\frac{1}{4\pi\epsilon_0}\frac{(-41*10^{-6})(0-12\hat{j})}{|0-12\hat{j}|^3}$

$\vec{E}=(5759.2\hat{i}+2559.6\hat{j})\,N/C$

Magnittude $E=6302\,N/C$

Direction $\theta=\tan^{-1}\frac{2559.6}{5759.2}=24\degree$ counterclockwise with the +ve X-axis.

b) Force on a charge $q=-2.10\,\mu C$ placed at origin is $\vec{F}=q\vec{E}$

$\vec{F}=(-2.10*10^{-6})(5759.2\hat{i}+2559.6\hat{j})=(-12.094*10^{-3}-5.369*10^{-3})\,N$

Magnitude $F=0.0132\,N$

Direction $\theta=180\degree+\tan^{-1}\left ( \frac{5.369}{12.094} \right )=204\degree$ counter clockwise with +ve X-axis.

Answer 2

Electric field at $\vec{r}$ at  due to a charge $q_1$ positioned at  $\vec{r}_1$  is $\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q_1(\vec{r}-\vec{r}_1)}{|\vec{r}-\vec{r}_1|^3}$

241 C
located at  $\vec{r}_2=8\hat{i}\,m$

$q_3=-41\,\mu C$ located at  $\vec{r}_3=12\hat{j}\,m$

$\vec{r}=0$ (origin)

a) Electric field at origin is  $\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q_2(\vec{r}-\vec{r}_2)}{|\vec{r}-\vec{r}_2|^3}+\frac{1}{4\pi\epsilon_0}\frac{q_3(\vec{r}-\vec{r}_3)}{|\vec{r}-\vec{r}_3|^3}$

$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{(-41*10^{-6})(0-8\hat{i})}{|0-8\hat{i}|^3}+\frac{1}{4\pi\epsilon_0}\frac{(-41*10^{-6})(0-12\hat{j})}{|0-12\hat{j}|^3}$

$\vec{E}=(5759.2\hat{i}+2559.6\hat{j})\,N/C$

Magnittude $E=6302\,N/C$

Direction $\theta=\tan^{-1}\frac{2559.6}{5759.2}=24\degree$ counterclockwise with the +ve X-axis.

b) Force on a charge $q=-2.10\,\mu C$ placed at origin is $\vec{F}=q\vec{E}$

$\vec{F}=(-2.10*10^{-6})(5759.2\hat{i}+2559.6\hat{j})=(-12.094*10^{-3}-5.369*10^{-3})\,N$

Magnitude $F=0.0132\,N$

Direction $\theta=180\degree+\tan^{-1}\left ( \frac{5.369}{12.094} \right )=204\degree$ counter clockwise with +ve X-axis.

Answer 3

Electric field at $\vec{r}$ at  due to a charge $q_1$ positioned at  $\vec{r}_1$  is $\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q_1(\vec{r}-\vec{r}_1)}{|\vec{r}-\vec{r}_1|^3}$

241 C
located at  $\vec{r}_2=8\hat{i}\,m$

$q_3=-41\,\mu C$ located at  $\vec{r}_3=12\hat{j}\,m$

$\vec{r}=0$ (origin)

a) Electric field at origin is  $\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q_2(\vec{r}-\vec{r}_2)}{|\vec{r}-\vec{r}_2|^3}+\frac{1}{4\pi\epsilon_0}\frac{q_3(\vec{r}-\vec{r}_3)}{|\vec{r}-\vec{r}_3|^3}$

$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{(-41*10^{-6})(0-8\hat{i})}{|0-8\hat{i}|^3}+\frac{1}{4\pi\epsilon_0}\frac{(-41*10^{-6})(0-12\hat{j})}{|0-12\hat{j}|^3}$

$\vec{E}=(5759.2\hat{i}+2559.6\hat{j})\,N/C$

Magnittude $E=6302\,N/C$

Direction $\theta=\tan^{-1}\frac{2559.6}{5759.2}=24\degree$ counterclockwise with the +ve X-axis.

b) Force on a charge $q=-2.10\,\mu C$ placed at origin is $\vec{F}=q\vec{E}$

$\vec{F}=(-2.10*10^{-6})(5759.2\hat{i}+2559.6\hat{j})=(-12.094*10^{-3}-5.369*10^{-3})\,N$

Magnitude $F=0.0132\,N$

Direction $\theta=180\degree+\tan^{-1}\left ( \frac{5.369}{12.094} \right )=204\degree$ counter clockwise with +ve X-axis.

Answer 4

Electric field at $\vec{r}$ at  due to a charge $q_1$ positioned at  $\vec{r}_1$  is $\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q_1(\vec{r}-\vec{r}_1)}{|\vec{r}-\vec{r}_1|^3}$

241 C
located at  $\vec{r}_2=8\hat{i}\,m$

$q_3=-41\,\mu C$ located at  $\vec{r}_3=12\hat{j}\,m$

$\vec{r}=0$ (origin)

a) Electric field at origin is  $\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{q_2(\vec{r}-\vec{r}_2)}{|\vec{r}-\vec{r}_2|^3}+\frac{1}{4\pi\epsilon_0}\frac{q_3(\vec{r}-\vec{r}_3)}{|\vec{r}-\vec{r}_3|^3}$

$\vec{E}=\frac{1}{4\pi\epsilon_0}\frac{(-41*10^{-6})(0-8\hat{i})}{|0-8\hat{i}|^3}+\frac{1}{4\pi\epsilon_0}\frac{(-41*10^{-6})(0-12\hat{j})}{|0-12\hat{j}|^3}$

$\vec{E}=(5759.2\hat{i}+2559.6\hat{j})\,N/C$

Magnittude $E=6302\,N/C$

Direction $\theta=\tan^{-1}\frac{2559.6}{5759.2}=24\degree$ counterclockwise with the +ve X-axis.

b) Force on a charge $q=-2.10\,\mu C$ placed at origin is $\vec{F}=q\vec{E}$

$\vec{F}=(-2.10*10^{-6})(5759.2\hat{i}+2559.6\hat{j})=(-12.094*10^{-3}-5.369*10^{-3})\,N$

Magnitude $F=0.0132\,N$

Direction $\theta=180\degree+\tan^{-1}\left ( \frac{5.369}{12.094} \right )=204\degree$ counter clockwise with +ve X-axis.