Q. 185.0( 2 Votes )

# A charged particle having a charge of –2.0 × 10–6 C is placed close to a nonconducting plate having a surface charge density 4.0 × 10–6 C m–2. Find the force of attraction between the particle and the plate.

Given:

Charge of the particle= -2.0× 10-6C=q

Surface charge density=4.0× 10-6C/m2

The electric field due to a plane thin sheet of charge density σ is given by

Proof:

To calculate the electric field at P we choose a cylindrical Gaussian surface as shown in the fig. in which the cross section A and A’ are at equal distance from the plane.

The electric field at all points of A have equal magnitude E. and direction along positive normal. The flux of electric field through A is given by

Since A and A’ are at equal distance from sheet the electric field at any point of A’ is also equal to E and flux of electric field through A’ is also given by

E.ΔS

At the points on curved surface the field and area make an angle of 90° with each other and hence

The total flux through the closed surface is given by

..,(i)

The area of sheet enclosed in the cylinder is given by ΔS

So the charge contained in the cylinder is given by

…(ii)

We know that,

By Gauss’s law, flux of net electric field (E ) through a closed surface S equals the net charge enclosed by the surface (qin) divided by ϵ0

Using gauss law and eqns(i) and (ii)

..(iii)

Now,

Force F on a charge particle of charge q in presence of electric field E is given by

Using eqn(iii),we get

Putting the values of σ and q we get

N

N(-ve sign indicates that the force is attractive in nature)

Therefore force of attraction between the particle and the plate is given by 0.45N

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