# The electric field in a region is given

Question:

The electric field in a region is given

$\overrightarrow{\mathrm{E}}=\left(\frac{3}{5} \mathrm{E}_{0} \hat{\mathrm{i}}+\frac{4}{5} \mathrm{E}_{0} \hat{\mathrm{j}}\right) \frac{\mathrm{N}}{\mathrm{C}}$. The ratio of flux of

reported field through the rectangular surface of area $0.2 \mathrm{~m}^{2}$ (parallel to $\mathrm{y}-\mathrm{z}$ plane) to that of the surface of area $0.3 \mathrm{~m}^{2}$ (parallel to $x-z$ plane) is $a: b$, where $a=$_________.

[Here $\hat{\mathrm{i}}, \hat{\mathrm{j}}$ and $\hat{\mathrm{k}}$ are unit vectors along $x, y$ and z-axes respectively]

Solution:

$\overrightarrow{\mathrm{E}}=\left(\frac{3 \mathrm{E}_{0}}{5} \hat{\mathrm{i}}+\frac{4 \mathrm{E}_{0}}{5} \hat{\mathrm{j}}\right) \frac{N}{\mathrm{C}}$

$\mathrm{A}_{1}=0.2 \mathrm{~m}^{2}$ [parallel to $\mathrm{y}-\mathrm{z}$ plane]

$=\overrightarrow{\mathrm{A}}_{1}=0.2 \mathrm{~m}^{2} \hat{\mathrm{i}}$

$\mathrm{A}_{2}=0.3 \mathrm{~m}^{2}$ [parallel to $\mathrm{x}-\mathrm{z}$ plane]

$\overrightarrow{\mathrm{A}}_{2}=0.3 \mathrm{~m}^{2} \hat{\mathrm{j}}$

Now $\phi_{\mathrm{a}}=\left[\frac{3 \mathrm{E}_{0}}{5} \hat{\mathrm{i}}+\frac{4 \mathrm{E}_{0}}{5} \hat{\mathrm{j}}\right] \cdot[0.2 \hat{\mathrm{i}}]=\frac{3 \times 0.2}{5} \mathrm{E}_{0}$

$\& \quad \phi_{\mathrm{b}}=\left[\frac{3 \mathrm{E}_{0}}{5} \hat{\mathrm{i}}+\frac{4 \mathrm{E}_{0}}{5} \hat{\mathrm{j}}\right] \cdot[0.3 \hat{\mathrm{j}}]=\frac{4 \times 0.3}{5} \mathrm{E}_{0}$

Now $\frac{\phi_{\mathrm{a}}}{\phi_{\mathrm{b}}}=\frac{0.6}{1.2}=\frac{1}{2}=\frac{\mathrm{a}}{\mathrm{b}}$

$\Rightarrow a: b=1: 2$

$\Rightarrow a=1$