Question

A charged particle having a charge of - 31C is placed close to a sheet of charge having a
charge density 5 * 10 Cm The force between the particle and sheet of charge is
1) 84 7 N (attraction)
2) 84.7 N (repulsion)
3) 0.847 N (attractive)
4) 0.847 N (repulsion)​

Answer 1

Answer:

Explanation:

The electric field due to charge plate with charge density (σ) is given by,

E = σ /2ϵ₀

Now the force between charge plate and charge q is,

F = qE

substituting the value of E in the equation,

F = qσ/2ϵ₀

Here,

• q = -31 C

• σ = 5×10¹⁰ C m

• ϵ₀ = 8.85 × 10⁻¹² F m⁻1

∴ F = (-31) × {(5×10¹⁰)/(2 × 8.85 × 10⁻¹²)}

⇒ F = -875.706 N

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