Q. 234.0( 4 Votes )

Two conducting plates X and Y, each having large surface area A (on one side), are placed parallel to each other as shown in figure. The plate X is given a charge Q whereas the other is neutral. Find:

(a) the surface charge density at the inner surface of the plate X,

(b) the electric field at a point to the left of the plates

(c) the electric field at a point in between the plates and

(d) the electric field at a point to the right of the plates



Answer :


Given:


Charge on plate X=Q


Charge on plate Y=zero


Consider the gaussian surface as shown in fig.


Two faces of this closed surface lie completely inside the conductor where the electric field is zero. The other parts of closed surface which are outside are parallel to electric field and hence flux on these parts is zero. The total flux of electric field through this closed surface is zero. So, from gauss’s law total charge inside the closed surface be zero. The charge on inner surface of X should be equal and opposite to charge on inner surface of Y.



To find the value of q consider electric field at point P


We know that electric field due to a thin plate of charge Q is given by



Where A=area of plate


Electric field at P


Due to charge Q-q=(towards right)


Due to charge q= (towards left)


Due to charge -q =(towards right)


Due to charge q= (towards left)


The net electric field at P due to all four charged surfaces is



Which is zero as P is inside conductor




The final charge distribution is as shown in fig



(a)surface charge density at inner surface of plate X is given by




Therefore surface charge density at inner surface of plate X is given by Q/2A


Consider Right direction as negative and left as positive


(b) at left of the plates


electric field due to,


outer surface of plate X=(towards left)


inner surface of plate X=(towards left)


inner surface of plate Y =(towards right)


outer surface of plate Y =(towards left)


net electric field at left of plates is given by



Therefore net electric field at a point to left to the plates is given by Q/2Aϵ0 towards left


(c)between the plates


electric field due to,


outer surface of plate X=(towards right)


inner surface of plate X=(towards right)


inner surface of plate Y =(towards right)


outer surface of plate Y =(towards left)


net electric field at a point between the plates is given by



Therefore net electric field at a point between the plates is given by Q/2Aϵ0 towards right


(d) to the right of the plates


electric field due to,


outer surface of plate X=(towards right)


inner surface of plate X=(towards right)


inner surface of plate Y =(towards left)


outer surface of plate Y =(towards right)


net electric field at a point between the plates is given by



Therefore net electric field at a point to the right of the plates is given by Q/2Aϵ0 towards right


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