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Grade: upto college level
        A charged particle having a charge of – 2.0 × 10^-6 is placed closed to a non-conducting plate having a surface charge density 4.0 × 10^-6 C/m^2. Find the force of attraction between the particle and the plate.
4 years ago

Answers : (1)

Aditi Chauhan
askIITians Faculty
396 Points
							Q = –2.0 × 10–6C Surface charge density = 4 × 10–6 C/m2
We know □(→┬E ) due to a charge conducting sheet = σ/2ε base 0
Again Force of attraction between particle & plate
= Eq = σ/2ε base 0 * q = 4 * 10^-6 * 2 * 10^-6/2 * 8 * 10^-12 = 0.452N

						
4 years ago
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