Physics, asked by sameerfea3988, 11 months ago

A charged particle having a charge of - 2.0 x 10^-6C is placed close to a nonconducting plate having a surface

Answers

Answered by jadondeeksha77

Explanation:

Q = –2.0 × 10–6C Surface charge density = 4 × 10–6C/m2 We know vector E due to a charge conducting sheet = σ/2ε0 Again Force of attraction between particcles

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