Question

A charged particle with a charge of −2⋅0 × 10−6 C is placed close to a non-conducting plate with a surface charge density of 4·0×10-6 C m-2. Find the force of attraction between the particle and the plate.

Answer 1

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Answer 2

The force of attraction between the plate and the particle is 0.45N.

Explanation:

Note, q = −2$\times$10−6C, σ = 4$\times$10−6Cm−2q = -2$\times$10-6C, σ = 4 $\times$ 10-6 Cm-2

The field of attraction between the charged particle and the plate,

F = qE

F = q $\times$ σ2∈0

F = (2.0$\times$10−6) $\times$ (8.85$\times$10−12) $\times$ (4.0$\times$10−6)2

F = 0.45 N

From the above calculations, it is provided that the attraction force between the particle and the plate is 0.45 N.