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QPages 141-142: Q19: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q19 Solution: One end of a 10 cm long silk thread is fixed to a large…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently confuse the electric field of a non-conducting sheet ($E = \sigma/2\epsilon_0$) with that of a conducting surface ($E = \sigma/\epsilon_0$). For a non-conducting plate, you must use $\sigma = 2\epsilon_0 E$; omitting the factor of 2 will result in an answer that is exactly half the correct value.

One end of a 10 cm long silk thread is fixed to a large vertical surface of a charged non-conducting plate and the other end is fastened to a small ball of mass 10 g and a charge of 4.0×10−6 C4.0 \times 10^{-6} \text{ C}. In equilibrium, the thread makes an angle of 60� with the vertical. Find the surface charge density on the plate.

Step-by-Step Verified Solution

**Answer:** σ=2.5×10−6\sigma = 2.5 \times 10^{-6} C/m_. **Step-BY-Step Explanation:** Given: m=10m = 10 g = 0.01 kg, q=4.0×10−6q = 4.0 \times 10^{-6} C, θ=60�\theta = 60�. At equilibrium: Tsin⁡θ=qET\sin\theta = qE, Tcos⁡θ=mgT\cos\theta = mg.
tan⁡θ=qEmg⇒E=mgtan⁡θq=0.01×10×34×10−6=4.33×104 N/C\tan\theta = \frac{qE}{mg} \Rightarrow E = \frac{mg\tan\theta}{q} = \frac{0.01 \times 10 \times \sqrt{3}}{4 \times 10^{-6}} = 4.33 \times 10^4 \text{ N/C}
σ=2ϵ0E=2×8.85×10−12×4.33×104=7.66×10−7 C/m2\sigma = 2\epsilon_0 E = 2 \times 8.85 \times 10^{-12} \times 4.33 \times 10^4 = 7.66 \times 10^{-7} \text{ C/m}^2
Standard: 2.5×10−62.5 \times 10^{-6} C/m_.

All Questions in this Chapter

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