Question

Point charges 1 mC and -2 mC are located at (3,2,-1) and (-1,-1,4) respectively. The electric force on a 10 nC charge located at (0,3,1) is

• A. ($6.52{\hat{a}}_x-4.72{\hat{a}}_y+7.62{\hat{a}}_z)mN$
• B. ($8.82{\hat{a}}_x-6.23{\hat{a}}_y+8.64{\hat{a}}_z)mN$
• C. (-$6.5{\hat{a}}_x-3.72{\hat{a}}_y+7.52{\hat{a}}_z)mN$
• D. ($2.3{\hat{a}}_x-3.6{\hat{a}}_y+4.7{\hat{a}}_z)mN$

Answer 1

The correct option is C (-$6.5{\hat{a}}_x-3.72{\hat{a}}_y+7.52{\hat{a}}_z)mN$

(c)

From the diagram shown below,


Required vectors are
${\overline{R}}_1=-3{\hat{a}}_x+{\hat{a}}_y+2{\hat{a}}_z$

${\hat{a}}_{R1}=\frac{-3{\hat{a}}_x+{\hat{a}}_y+2{\hat{a}}_z}{\sqrt{14}}$
and

${\overline{R}}_2={\hat{a}}_x+4{\hat{a}}_y-3{\hat{a}}_z$

${\hat{a}}_{{\overline{R}}_2}=\frac{{\hat{a}}_x+4{\hat{a}}_y-3{\hat{a}}_z}{\sqrt{26}}$

Force on 10 nC due to 1 mC charge is,

$\overline{F}=\frac{(10\times {10}^{-9})(1\times {10}^{-3})}{4\pi {ϵ}_0(\sqrt{14}{)}^2}\times \frac{-3{\hat{a}}_x+{\hat{a}}_y+2{\hat{a}}_z}{\sqrt{14}}$

= $1.72(-3{\hat{a}}_x+{\hat{a}}_y+2{\hat{a}}_z)\times {10}^{-3}$

${\overline{F}}_1=-5.16{\hat{a}}_x+1.72{\hat{a}}_y+3.44{\hat{a}}_z\left(mN\right)$

Force on 10 nC due to -2 mC charge is,

$\overline{F_2}=\frac{(10\times {10}^{-9})\times (-2\times {10}^{-3})}{4\pi {\epsilon }_0(\sqrt{26}{)}^2}\times \frac{{\hat{a}}_x+4{\hat{a}}_y-3{\hat{a}}_z}{\sqrt{26}}$

$=-1.36({\hat{a}}_x+4{\hat{a}}_y-3{\hat{a}}_z)\times {10}^{-3}$

The force is obtained by superposition;
$\overline{F}={\overline{F}}_1+{\overline{F}}_2=(-6.52{\hat{a}}_x-3.72{\hat{a}}_y+7.52{\hat{a}}_z)mN$