Concept on electric field:
1. Field of two oppositely charged sheets:
Two infinite plane sheets each of surface charge density σ and –σ are placed parallel to each other at a small distance d. Electric field for sheet 1 and 2 are E1 = E2 = 
.
Feld at point A is 
+ 
0
Feld at point B is 
+ 
Feld at point C is 
+ 0.
2. Field of two equally charged sheets:
Two infinite plane sheets each of surface charge density σ are placed parallel to each other at a small distance d. Electric field for sheet 1 and 2 are E1 = E2 = .
Feld at point A is +
Feld at point B is + 0
Feld at point C is + .
Concept on principle of super position of electric field:
1. An infinite plane sheet with of surface charge density σ has a hole of radius R in it. Calculate the electric field at point P at a distance x on the axis of the hole from it’s centre.
The infinite charged sheet with a circular hole can be considered as the superposition of an infinite sheet of charge density σ and charged disc of radius R with charge density -σ.
The electric field at point P due to infinite plane sheet is 
= 
î
The electric field at point P due to charged disc of radius R is 
= 
(-î)
So, net electric field at point P due to the hollow charged sheet is 
= + = 
.
If an electron is released from point P (in previous concept) then what is the speed of it at the centre of the hole?
The force acing on the electron is 
= – q = – 
Or, mv 
= – 
Or, mvdv = – 
When the electron is at a distance x (point P) the speed is 0 and when it is at the centre of the hole (x = 0) then speed is v.
Or, m
vdv = – 
dx
[Let, R2 + x2 = z2 or, 2xdx = 2zdz or, xdx = zdz and the limits of z is 
to R. Then, 
= z]
Or, 
v = 
.
2. A sphere of radius R has a uniform volume charge density 
. A sphere of cavity of radius b is at a distance a from the centre of the sphere is removed. Find the electric field at any point P inside the cavity.
The electric field inside the cavity is the superposition of the electric field due to the original uncut sphere of charge density , plus the electric field due to the sphere of the size of the cavity of charge density – .
The electric field at point P at a distance r from O due to the original uncut sphere is = 
r̂ = 
.
The electric field at point P at a distance c from the centre of the cavity with charge density – is = 
ĉ = 
.
From figure we get, 
= 
+ 
Or, = –
So, = 
.
The net electric field at point P is = + = + = 
.
Speed of charge in electric field:
An electron of charge e and mass m is projected with an initial velocity at an angle 
to the horizontal from the lower plate of a parallel-plate capacitor as shown in figure. The plates are sufficiently long and have a separation of d. Find the maximum value of the velocity of the particle so that it does not hit the upper plate. The electric field E between the plates is directed upward.
Let the maximum value of the initial velocity of the electron is u. Resolving the velocity of the particle parallel and perpendicular to the plate, we get 
= u cos and 
= u sin . Force on electron in the downward direction and normal to the plate is qE. Therefore, downward acceleration a = 
.
The particle will not hit the upper plate if the velocity component normal to the plate becomes zero before reaching it, i.e. 0 = 
– 2ad
Or, 0 = u2sin2 – 
Or, u = 
.
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