Question

In the figure shown, the particles have charges $q_1$ = $-q_2$ = 300 nC and $q_3$ = $-q_4$ =200 nC, and distance a = 5.0 cm. What are the (A) magnitude and (B) angle (relative to the +x direction) of the net force on particle 3?

• A. 0.4 $\hat{i}$ – 0.188 $\hat{j}$ N
• B. 0.5 $\hat{i}$ – 0.188 $\hat{j}$ N
• C. 0.4 $\hat{i}$ + 0.188 $\hat{j}$ N
• D. 0.5 $\hat{i}$ – 0.188 $\hat{j}$ N

Answer 1

The correct option is A

0.4 $\hat{i}$ – 0.188 $\hat{j}$ N

To find net force on $q_3$, we use principle of superposition. We just have to find vector sums of forces due to each charge on $q_3$. Let their magnitudes be $F_{13},F_{23},F43$

$F_{13}=\frac{K{q}_1{q}_3}{a^2}$

$=\frac{9\times {10}^9\times 300\times {10}^{-9}\times 200\times {10}^{-9}}{(5\times {10}^{-2}{)}^2}$

$=2.16\times {10}^{-1}N$

= 0.216 N

${\overrightarrow{F}}_{13}$ = –0.216 $\hat{j}$ N (since it’s a repulsive force and acts in the –ve y direction (radially)

$F_{23}=\frac{K{q}^2{q}^3}{(5\sqrt{2}\times a^{-2}{)}^2}$ (we need not take ‘–‘sign on $q_2$. We can figure out direction at the end)

$=\frac{9\times {10}^9\times 300\times {10}^{-9}\times 200\times {10}^{-9}}{(5\times {10}^{-2}{)}^2}$ (because length of a square’s diagonal is $\sqrt{2}\times$ side)

= 0.108 N

${\overrightarrow{F}}_{23}$ is attractive and makes ${45}^{\circ }$ with +X direction

${\overrightarrow{F}}_{23}$ = 0.108 sin 45 $\hat{i}$ + 0.108 cos 45 $\hat{j}$

= 0.764 $\hat{i}$ + 0.764 $\hat{j}$

$F_{43}=\frac{K{q}^4{q}^3}{a^2}$

$=\frac{9\times {10}^9\times 300\times {10}^{-9}\times 200\times {10}^{-9}}{(5\times {10}^{-2}{)}^2}$

= 0.324 N

${\overrightarrow{F}}_{23}$ = is attractive and is in +X direction

${\overrightarrow{F}}_{23}$ = 0.324 $\hat{i}$ N

Now total force on q3, ${\overrightarrow{F}}_{43}$ = ${\overrightarrow{F}}_{13}+{\overrightarrow{F}}_{23}+{\overrightarrow{F}}_{43}$

= –0.216 $\hat{j}$ + 0.0764 $\hat{i}$+ 0.764 $\hat{j}$+ 0.324 $\hat{i}$

= 0.4004 $\hat{i}$ – 0.1876 $\hat{j}$ N

${\overrightarrow{F}}_3$= 0.4 $\hat{i}$ – 0.188 $\hat{j}$ N