Question

Point charges of $+2\mu C,+4\mu C,-6\mu C$ and $+8\mu C$ are placed at the corners A,B,C and respectively of a square ABCD of side 0.2 m. Calculate the resultant force on the charge at D.

Answer 1

Given four charges placed at corners $A$, $B$, $C$ and $D$ of square of side $0.2m$.

$AD=CD=0.2m$

$BD=\sqrt{(o.2{)}^2+(0.2{)}^2}$

$=0.283m$

So, force on $+8\mu C$ due to charge at $A$

$F_1=\frac{1}{4\pi {\epsilon }_0}\frac{2\times 8\times {10}^{-12}}{(0.2{)}^2}$

$=9\times {10}^9\times 4\times {10}^{-10}$

$=3.6N$ along $AD$

Force on $+8\mu C$ due to charge at $B$

$F_2=\frac{1}{4\pi {\epsilon }_0}\frac{32\times 8\times {10}^{-12}}{(0.2{)}^2}$

$=9\times {10}^9\times 408.16\times {10}^{-12}$

$=3673.47\times {10}^{-3}$

$=3.673N$ along $BD$

Force on $+8\mu C$ due to charge at $C$

$F_3=9\times {10}^9\times \frac{48\times {10}^{-12}}{(0.2{)}^2}$

$=10800\times {10}^{-3}$

$=10.8N$ along $AD$

Resultant of $F_1$ and $F_2$ is $\sqrt{(3.6{)}^2+(10.8{)}^2}$

$F^{\prime }=\sqrt{12.96+116.64}$

$=\sqrt{129.6}=11.38N$

Now, net force $=$ Resultant of $F^{\prime }$ and $F_2$

$=\sqrt{(F^{\prime }{)}^2+(F_2{)}^2}$

$=\sqrt{(11.38{)}^2+(3.673{)}^2}$ along $F_1$

$=\sqrt{129.50+1.491}$

$=\sqrt{142.99}=11.958N$