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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 × 10-6Cm0-2. Find the force of attraction between the particle and the plate.
Solution
The electric field due to a conducting thin sheet,
`"E" = sigma/( 2 ∈ _0)`
The magnitude of attractive force between the particle and the plate,
F =qE
`"F" = (q xx sigma)/(2∈_0)`
`"F" = ((2.0 xx 10^-6 ) xx ( 4.0 xx 10^-6))/(2 xx (8.55 xx 10^-12))`
F = 0.45 N
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