Question

The electric field is $\vec{E} = \left( \frac{3}{5} E_0 \hat{i} + \frac{4}{5} E_0 \hat{j} \right)$. The ratio of flux through surface of area 0.2 m² (parallel to y-z plane) to that of area 0.3 m² (parallel to x-z plane) is a : b, where a = _________.

Hint:

Remember: A surface parallel to the $y-z$ plane has its normal (area vector) along the $x$-axis ($\hat{i}$). Only the $x$-component of the electric field will contribute to flux through it.

Answer 1

Correct Answer: 1

Step 1: Flux $\Phi = \vec{E} \cdot \vec{A}$.
Step 2: Surface 1 (parallel to y-z plane): Area vector $\vec{A}_1 = 0.2 \, \hat{i}$. $\Phi_1 = (\frac{3}{5}E_0 \hat{i} + \frac{4}{5}E_0 \hat{j}) \cdot (0.2 \hat{i}) = \frac{3}{5} E_0 (0.2) = 0.12 E_0$.
Step 3: Surface 2 (parallel to x-z plane): Area vector $\vec{A}_2 = 0.3 \, \hat{j}$. $\Phi_2 = (\frac{3}{5}E_0 \hat{i} + \frac{4}{5}E_0 \hat{j}) \cdot (0.3 \hat{j}) = \frac{4}{5} E_0 (0.3) = 0.24 E_0$.
Step 4: Ratio $\frac{\Phi_1}{\Phi_2} = \frac{0.12}{0.24} = \frac{1}{2}$.
Step 5: Ratio is $1:2$, so $a=1$.