Question

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

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

Answer 1

Answer: (A)

$A D=B C=\sqrt{(5)^{2}+(5)^{2}}=\sqrt{25+25}=\sqrt{2} \times 5 cm$

$\Rightarrow A O=B O=C O=O D=\frac{5 \sqrt{2}}{2} cm$
The electric field,
$E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}}$
So, $E_{A}=\frac{9 \times 10^{9} \times 10 \times 10^{-9} \times 4}{25 \times 2}$
$=7.2\, N \,C^{-1}$ along $O D$
$E_{B}=\frac{9 \times 10^{9} \times 2 \times 10 \times 10^{-9} \times 4}{25 \times 2}$
$=14.4\, N\, C^{-1} $ along $O B$
$E_{C}=\frac{9 \times 10^{9} \times 10 \times 10^{-9} \times 4}{25 \times 2}$
$=7.2$ along $OC$
$E_{D}=\frac{9 \times 10^{9} \times 2 \times 10 \times 10^{-9} \times 4}{25 \times 2}$
$=14.4$ along $O A$
Resultant of $E_{A}$ and $E_{D'} E_{1}=(14.4-7.2)$
$=7.2\, N\, C^{-1}$ along $O A$
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 $O A$ upward.