Question

2. Charge A, acharge of +2.00 μC, is placed at the origin. Ch -3.00pc, arge B, a charge of placed on the x-axis at x = 1.00 m. Charge C, a charge of-4.00 μ C, is placed on the x-axis at 1.00 m. What is the total electric field felt by Charge A? (That is, how many Newtons of force, and does it point in the +x direction or in the -x direction? (continued on the next page)

Answer 1

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

Here charges and their locations are

$q_A=+2.00\,\mu{C}$ and $\vec{r}_A=0$

$q_B=-3.00\,\mu{C}$ and $\vec{r}_B=1.00\,\hat{i}\,m$

$q_C=-4.00\,\mu{C}$ and $\vec{r}_C=-1.00\,\hat{i}\,m$

Net electric field at the location of A is $\vec{E}_{A,net}=\frac{q_B}{4\pi\epsilon_0}\frac{(\vec{r}_A-\vec{r}_B)}{|\vec{r}_A-\vec{r}_B|^3}+\frac{q_C}{4\pi\epsilon_0}\frac{(\vec{r}_A-\vec{r}_C)}{|\vec{r}_A-\vec{r}_C|^3}$

$\vec{E}_{A,net}=\frac{-3.00*10^{-6}}{4\pi\epsilon_0}\frac{(0-1.00\,\hat{i})}{|0-1.00\,\hat{i}|^3}+\frac{-4.00*10^{-6}}{4\pi\epsilon_0}\frac{(0+1.00\,\hat{i})}{|0+1.00\,\hat{i}|^3}$

$\vec{E}_{A,net}=\frac{3.00*10^{-6}}{4\pi\epsilon_0}\frac{(1.00\,\hat{i})}{1.00}+\frac{-4.00*10^{-6}}{4\pi\epsilon_0}\frac{(1.00\,\hat{i})}{1.00}$

$\vec{E}_{A,net}=(8.9875*10^{9})(3.00*10^{-6}\,\hat{i}-4.00*10^{-6}\,\hat{i})=-8987.5\,\hat{i}$

Electric Force on charge A is $\vec{F}_{A,net}=q_A\vec{E}_{A,net}=(2.00*10^{-6})(-8987.5\,\hat{i})=-0.0180\,\hat{i}\,N$

Magnitude of force on charge A is $F_{A,net}=0.0180=1.80*10^{-2}\,\,N$

Force on charge A and field at location of A are along negative X direction.