Question

Question 5 In the figure, the particles have charges q1 =-q2 = 670 nC and q3 =-q: 98 nC, and distance a = 5.6 cm. What are the (a) x and (b) y components of the net electrostatic force on particle 3? 3 (a) Number (b) Number Click if you would like to Show Work for this question: Units Units Open Show Work

Answer 1

Charge $q_1=670\,nC$ is located at $\vec{r}_1=a\hat{j}$

Charge $q_2=-670\,nC$ is located at $\vec{r}_2=a\hat{i}+a\hat{j}$

Charge $q_3=98\,nC$ is located at $\vec{r}_1=0\hat{i}+0\hat{j}$ ( at origin)

Charge $q_4=-98\,nC$ is located at $\vec{r}_4=a\hat{i}$

($\hat{i},\hat{j}$are unit vectors along positive X and positive Y axes respectively)

$(k=8.99*10^9\,N.m^2/C^2)$

$a=5.6\,cm=0.056\,m$

Net electrostatic force on particle 3 is $\vec{F}_3=\frac{kq_3q_1(\vec{r}_3-\vec{r}_1)}{|\vec{r}_3-\vec{r}_1|^3}+\frac{kq_3q_2(\vec{r}_3-\vec{r}_2)}{|\vec{r}_3-\vec{r}_2|^3}+\frac{kq_3q_4(\vec{r}_3-\vec{r}_4)}{|\vec{r}_3-\vec{r}_4|^3}$

$\vec{F}_3=kq_3\left [ \frac{q_1(\vec{r}_3-\vec{r}_1)}{|\vec{r}_3-\vec{r}_1|^3}+\frac{q_2(\vec{r}_3-\vec{r}_2)}{|\vec{r}_3-\vec{r}_2|^3}+\frac{q_4(\vec{r}_3-\vec{r}_4)}{|\vec{r}_3-\vec{r}_4|^3} \right ]$

$\vec{F}_3=kq_3\left [ \frac{q_1(-a\hat{j})}{|-a\hat{j}|^3}+\frac{q_2(-a\hat{i}-a\hat{j})}{|-a\hat{i}-a\hat{j}|^3}+\frac{q_4(-a\hat{i})}{|-a\hat{i}|^3} \right ]$

$\vec{F}_3=\frac{kq_3}{a^2}\left [ q_1(-\hat{j})+\frac{q_2(-\hat{i}-\hat{j})}{2\sqrt{2}}+q_4(-\hat{i}) \right ]$

$\vec{F}_3=\frac{8.99*10^9*98*10^{-9}}{(0.056)^2}\left [ 670*10^{-9}(-\hat{j})+\frac{(-670*10^{-9})(-\hat{i}-\hat{j})}{2\sqrt{2}}+(-98*10^{-9})(-\hat{i}) \right ]$

$\vec{F}_3=\frac{8.99*10^9*98*10^{-9}}{(0.056)^2}(334.9*10^{-9}\,\hat{i}-433.1*10^{-9}\,\hat{j})$

$\vec{F}_3=(0.094\hat{i}-0.122\hat{j})$

a) X-component of electrostatic force on particle 3 is $0.094\,N$

b) Y-component of electrostatic force on particle 3 is $-0.122\,N$