Question

18. A charged particle having a charge of - 2:0 x 10°C is
placed close to a nonconducting plate having a surface
charge density 4:0 x 10-C m? Find the force of
attraction between the particle and the plate.

Answer 1

Answer:

The electric field due to charge plate with charge density σ is given by, E=

2ϵ

0

σ

Now the force between charge plate and charge q is F=qE=q

2ϵ

0

σ

or F=(−2×10

−6

)×

2×8.85×10

−12

4×10

−6

=−0.45N

(Negative sign indicates the force between them is attractive)