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QPages 141-142: Q18: HC Verma: Concepts of Physics - Volume 2 Gauss's Law Pages 141-142: Q18 Solution: A charged particle with a charge of is placed close to a…

A charged particle with a charge of −2.0×10−6 C-2.0 \times 10^{-6} \text{ C} is placed close to a non-conducting plate with a surface charge density of 4.0×10−6 C m−24.0 \times 10^{-6} \text{ C m}^{-2}. Find the force of attraction between the particle and the plate.

Step-by-Step Verified Solution

**Answer:** F=0.45F = 0.45 N. **Step-by-Step Explanation:** Given: q=−2.0×10−6q = -2.0 \times 10^{-6} C, σ=4.0×10−6\sigma = 4.0 \times 10^{-6} C/m_. Field due to sheet: E=σ2ϵ0=4.0×10−62×8.85×10−12=2.26×105E = \frac{\sigma}{2\epsilon_0} = \frac{4.0 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = 2.26 \times 10^5 N/C.
F=qE=2.0×10−6×2.26×105=0.45 NF = qE = 2.0 \times 10^{-6} \times 2.26 \times 10^5 = 0.45 \text{ N}

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