Let, the midpoint of the line joining the two charges be O.
a) Electric field at O due to charge at A-EA=
Electric field at O due to charge at B-EB=
Resultant intensity -E =EA+ EB
=
b) If a test charge of magnitude is placed at the centre, then force experienced is
Given, q1= +0.2 C
q2= +0.4 C
distance between the charges-d=0.1 m
a)Electric field at the mid point between these two charges:
Electric field due to q1-E1=
/C
Electric field due to q2-E2=
/C
Resultant Electric field at mid-point-E=
Since the net electric field is acting in opposite direction we have E= 1440
= /C
b)
Let point P be on the line joining the charges such that it is 0.05m away from q2 and 0.15 m away from q1.
Electric field due to q1=
Electric field due to q2=/C
Since, Electric field is acting in the same direction
Resultant electric field intensity-E=
Consider a thin spherical shell of radius 'R' with centre O. Let, a charge +q be distributed uniformly on the surface of the shell.
According to gaussian theorem,
Surface area of the sphere = 4r2
Now, at a point on the surface area of the shell
Let, be the surface charge density on the shell,
then,
q = 4R2 .
Therefore,
which is the required electric field at the surface of the shell.