Q. 223.7( 3 Votes )

# Two large conducting plates are placed parallel to each other and the carry equal and opposite charges with surface density σ as shown in figure. Find the electric field

(a) at the left of the plates.

(b) in between the plates and

(c) at the right of the plates.

Answer :

Given:

Surface charge density of both plate=σ

Both plates carry equal and opposite charges

We know that,

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 (E⃗ ) through a closed surface S equals the net charge enclosed by the surface (q_{in}) divided by ϵ_{0}

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

Now

Magnitude of Electric field due to plate 1 is given by

Magnitude of Electric field due to plate 2 is given by

(a)at the left of the plates

Electric field due to plate 1 is in left direction(-ve) whereas electric field due to plate 2 is in right direction(+ve)

∴ net electric field

**Therefore net electric field due to both plates at the left of plates is zero**

(b)in between the plates,

The electric field due to plate 1 is in right direction(+ve) and electric field due to plate 2 is also in right direction(+ve)

∴ net electric field

**Therefore net electric field due to both plates in between the plates is given by σ/****ϵ**_{0}

(c)at the right of the plates

Electric field due to plate 1 is in right direction (+ve) whereas electric field due to plate 2 is in left direction (-ve)

∴ net electric field

**Therefore net electric field due to both plates in the right of both plates is zero**

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