Question

What is the direction of the electric field at the centre $O$ of the square in the figure shown below? Given that, $q=10nC$ and the side of the square is $5cm$.

• A. at $45^{\circ }$ to $OA$ upward
• B. at $135^{\circ }$ to $OA$ towards $BD$
• C. no direction hecause $F=0$
• D. None of the above

Answer 1

Answer: (A)

$AD=BC=\sqrt{(5{)}^2+(5{)}^2}=\sqrt{25+25}=\sqrt{2}\times 5cm$

$\Rightarrow AO=BO=CO=OD=\frac{5\sqrt{2}}{2}cm$
The electric field,
$E=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q}{r^2}$
So, $E_A=\frac{9\times 10^9\times 10\times 10^{-9}\times 4}{25\times 2}$
$=7.2N{C}^{-1}$ along $OD$
$E_B=\frac{9\times 10^9\times 2\times 10\times 10^{-9}\times 4}{25\times 2}$
$=14.4N{C}^{-1}$ along $OB$
$E_C=\frac{9\times 10^9\times 10\times 10^{-9}\times 4}{25\times 2}$
$=7.2$ along $OC$
$E_D=\frac{9\times 10^9\times 2\times 10\times 10^{-9}\times 4}{25\times 2}$
$=14.4$ along $OA$
Resultant of $E_A$ and $E_{D^{\prime }}{E}_1=(14.4-7.2)$
$=7.2N{C}^{-1}$ along $OA$
Since, $E_1$ and $E_2$ are perpendicular to each other.
$\therefore E=\sqrt{E_1{}^2+{E_2}^2}$ is along $45^{\circ }$ to $OA$ upward.