Answer to Question #295561 in Electric Circuits for John

Question

Consider two charges q1=38e and q2=−48e at positions (−22,−26,−14) and (46/√3, 20/√2, −35) respectively where all the coordinates are measured in the scale of 10^−9m or nano meters. If position vector of the charge q1 is r⃗ 1 and charge q2 is r⃗ 2.

find x,y and z components of vector for the following questions

a) Find the relative position vector that points from charge q1 to charge q2.

b) Calculate the force acting on q1 due to q2.

Now consider another charge q3=20e is in the xyz system positioned at (32 /√3, −16/√2, −15). Calculate the net force acting on q1 and q2.

c) Net Force on q1 due to other charges

d) Net Force on q2 due to other charges

e) Calculate the net electric field at the position of q3 due to q1 and q2 charge. Note that, you cannot calculate electric field of q3 at it's own position.

Say the charge q3 have mass of m=0.18gram. What will be it's acceleration given it's in the field of other charges.

f) Calculate the magnitude of the acceleration of q3 charge

Answer 1

a)

$r_{12}=r_2-r_1=(48 .56,40.14,-21)nm$

b)

$\begin{vmatrix} r_{12}\ \end{vmatrix}=\begin{vmatrix} r_2-r_1 \\ \end{vmatrix}$

$=\sqrt{(\frac{46}{\sqrt3}+22)^2+(\frac{20}{\sqrt2}+26)^2+(-21)^2}$

$=6. 641×10^{-8}m$

$\hat{r}_{12}=\frac{r_{12}}{|r_{12}|}=(0.7312,0.6044,-0.3162)$

$\vec{F}_{12}=\frac{kq_1q_2}{r^2}\>\>\hat{r}_2$

$=\frac{9×10^9×38×48×(1.6×10^{-19})^2}{(6.641×10^{-8})^2}$ $\begin{bmatrix} 0.7312,0.6044,-0.3162 \\ \end{bmatrix}$

$=(6.967,5.759,-3.013)×10^{-11}$

The force is in direction $r_{12}$

c)

$|r_{13}|=\begin{vmatrix} r_3 -r_1 \\ \end{vmatrix}$

$=\sqrt{40.48^2+14.69^2+(-1)^2}$

$=4.307×10^{-8}m$

$\begin{vmatrix} \hat{r}_{13} \\ \end{vmatrix}=\frac{r_{13}}{|r_{13}|}$ $=(0.9399,0.3411,-0.02322)$

$F_{13}=\frac{9×10^9×20×38(1.6×10^{-19})^2}{(4 .307×10^{-8})^2}$ $\begin{bmatrix} 0.9399,0.3411,-0.02322 \\ \end{bmatrix}$

$=(8.872,3.220,-0.2192)×10^{-11}N$

The force is in direction of $r_{13}$

d)

$\begin{vmatrix} r_{23}\\ \end{vmatrix}=\begin{vmatrix} r_3 -r_2 \\ \end{vmatrix}$

$=\sqrt{(-8.083^2)+(-25.46)^2+(20)^2}$

$=3.337×10^{-8}m$

$\begin{vmatrix} \hat{r}_{23} \\ \end{vmatrix}=\frac{r_{23}}{|r_{23}|}=(-0.2422,-0.7630,0.5994)$

$F_{23}=\frac{9×10^9×20×-48×(1.6×10^{-19})^2}{(3.337×10^{-8})^2}$ $\begin{bmatrix} -0.2422,-0.7630,0.5994 \\ \end{bmatrix}$

$=(4.811,15.16,-11.91)×10^{11}N$

The force is in direction of $r_{23}$

e)

$E=\frac{F}{q}$

$E_1$ on $q_3$ due to $q_1$

$E_1=\frac{F_{13}}{20×1.6×10^{-19}}$

$=(2.773,1.006,0.0685)×10^7$ $N/C$

$E_2$ on $q_3$ is due to $q_1$

$E_2=\frac{F_{23}}{20×1.6×10^{-19}}$

$=(1.503, 4.738, -3.722)×10^7\>N/C$

f)

$a=\frac{F_{net}}{Mass}$

$F_{net}=F_{13}+F_{23}$

$F_{net}=(8.872+4.811,3.220+15.16,-0.2192-11.91)×10^{-11}$

$a=(7.602,10.21,-6.738)×10^{-7}\>m/s^2$

$|a|=1.44×10^{-6}\>m/s^2$