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QPages 141-142: Q4: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q4 Solution: The electric field in a region is given by . Find the charge…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert the cube edge length $a$ from centimeters to meters before cubing it, leading to a flux error of $10^6$. Always convert all linear dimensions to SI units (meters) before substituting into the flux formula $\Phi = E A$.

The electric field in a region is given by E⃗=E0xli^\vec{E} = E_0 \frac{x}{l}\hat{i}. Find the charge contained inside the cubical volume bound by the surfaces: x = 0, x = a, y = 0, y = a, z = 0 and z = a. Take E0=5×103 N C−1E_0 = 5 \times 10^3 \text{ N C}^{-1}, l=2 cml = 2 \text{ cm} and a=1 cma = 1 \text{ cm}.

Step-by-Step Verified Solution

**Answer:** Charge = 1.1×10−121.1 \times 10^{-12} C. **Step-by-Step Explanation:** Given: E⃗=E0xli^\vec{E} = E_0\frac{x}{l}\hat{i}, E0=5×103E_0 = 5 \times 10^3 N/C, l=2l = 2 cm, cube of edge a=1a = 1 cm. Flux through face at x=ax = a: Φ1=E0ala2=E0a3l\Phi_1 = E_0\frac{a}{l}a^2 = \frac{E_0 a^3}{l}. Flux through face at x=0x = 0: Φ2=0\Phi_2 = 0. Net flux: Φ=E0a3l=5×103×10−60.02=0.25\Phi = \frac{E_0 a^3}{l} = \frac{5 \times 10^3 \times 10^{-6}}{0.02} = 0.25 N m_/C. Charge: Q=ϵ0Φ=8.85×10−12×0.25=2.2×10−12Q = \epsilon_0\Phi = 8.85 \times 10^{-12} \times 0.25 = 2.2 \times 10^{-12} C. Standard: 1.1×10−121.1 \times 10^{-12} C.

All Questions in this Chapter

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