Question

In the given arrangement of a charged square frame find electric field at centre. The linear charge density is as shown in figure.

• A. $\frac{\lambda \sqrt{2}}{4\pi {ϵ}_{0}l}[-2i+14j]$
• B. $\frac{\lambda \sqrt{2}}{4\pi {ϵ}_{0}l}[-2i+14j]$
• C. Zero
• D. None of these

Answer 1

The correct option is A $\frac{\lambda \sqrt{2}}{4\pi {ϵ}_{0}l}[-2i+14j]$

Electric field due to 1 is
$E_1=\frac{\lambda \sqrt{2}}{4\pi {ϵ}_{0}l}(cos{45}^{\circ }+cos{45}^{\circ })\hat{i}$
$\Rightarrow E_1=\frac{1}{4\pi {ϵ}_0}\left(\frac{2\sqrt{2\lambda }}{l}\right)\hat{i}$
Electric field due to 2 is $E_2=\frac{1}{4\pi {ϵ}_0}(-\frac{4\sqrt{2\lambda }}{l})\hat{j}$
Electric field due to 3 is $E_3=\frac{1}{4\pi {ϵ}_0}\left(\frac{6\sqrt{2\lambda }}{l}\right)\hat{j}$
Electric field due to 4 is $E_3=\frac{1}{4\pi {ϵ}_0}\left(\frac{8\sqrt{2\lambda }}{l}\right)\hat{j}$
$\Rightarrow \overrightarrow{E_{total}}=\overrightarrow{E_1}+\overrightarrow{E_2}+\overrightarrow{E_3}+\overrightarrow{E_4}$
$\Rightarrow \overrightarrow{E}=\frac{1}{4\pi ϵ}(\frac{2\sqrt{2}\lambda }{l}-\frac{4\sqrt{2}\lambda }{l})\hat{i}+\frac{1}{4\pi ϵ}(\frac{6\sqrt{2}\lambda }{l}+\frac{8\sqrt{2}\lambda }{l})\hat{j}$
$\overrightarrow{E}=\frac{\lambda \sqrt{2}}{4\pi {ϵ}_{0}l}(-2\hat{i}+14\hat{j})$