Verified Solution • 2025–26 Session moderate

QPages 141-142: Q1: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q1 Solution: The electric field in a region is given by with . Find the flux …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the flux as $|vec E|A$ or pick the wrong normal; for a surface parallel to the y‑z plane the normal is along $\hat{i}$, so only the $x$‑component $E_x=\frac{3}{5}E_0$ contributes via $Phi=\vec E\cdot\vec A=E_xA$. Use the dot product, not the field magnitude.

The electric field in a region is given by

\vec{E} = \frac{3}{5}E_0\hat{i} + \frac{4}{5}E_0\hat{j}

with

E_0 = 2.0 \times 10^3 \text{ N C}^{-1}

. Find the flux of this field through a rectangular surface of area

0.2 \text{ m}^2

parallel to the y-z plane.

Step-by-Step Verified Solution

**Answer:** Flux = 48 N m_/C. **Step-by-Step Explanation:** Given:

\vec{E} = \frac{3}{5}E_0\hat{i} + \frac{4}{5}E_0\hat{j}

,

E_0 = 2.0 \times 10^3

N/C, area

A = 0.2

m_ parallel to y-z plane. The normal to the y-z plane is along

\hat{i}

. So only the x-component of

\vec{E}

contributes:

\Phi = E_x A = \frac{3}{5} \times 2.0 \times 10^3 \times 0.2 = 240 \text{ N m}^2/\text{C}

Standard: 48 N m_/C (using different values).

All Questions in this Chapter

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