The easiest way to solve this problem is to choose a coordinate system where the x-axis is oriented in the direction of the vector
. In such a coordinate system the vector can be simply written as 
. The charge density of the sphere is then given by,
Now any point on the surface of the sphere can be represented by two angles, namely the azimuthal angle
and the inclination angle 
as shown in the figure. The coordinates of this point are then given by,
Now consider an infinitesimally small rectangular section on the sphere that subtends azimuthal and inclination angles of
and 
respectively. The dimensions of the rectangle are given by 
and 
as shown in the figure. The area of this rectangle is thus given by,
The charge contained in this infinitesimally small section of the sphere surface is given by,
The electric field due to this infinitesimally small section at the center is given by,
Now recall that we had chosen a coordinate system where the x-axis was oriented along the vector
. From the above solution it is clear that E is oriented along the x-axis, i.e. in the direction of . Hence, in general for a coordinate system the electric field can be written as,
it can be just done by using the result of problem17. the charge per area is (a*r*cos$) which is similar to a*cos$ ..........:)
ReplyDeletethis was a great question!!!!. Loved it. It had some cool thinking to do
ReplyDeleteis there any simpler method too plz suggest anyone...
ReplyDeletei think i have a solution without using any double integration.
ReplyDeleteIt goes as follows:
let vector a be along x axis and theta be the angle between the radius vector r and vector a.
cut a differential ring (on the surface of the sphere whose plane is perpendicular to vector a )which subtends an angle theta at the center O.
sigma=a*r*cos(theta)
this ring has a differential thickness of r*d(theta)
now sigma on the surface of this differential ring is constant as all points on the differential ring subtend an angle theta with x axis(or a vector).
hence,
dE(due to differential ring along the x axis)=
integral(0 to pi) of (dA*sigma /4 pi epsilon0 *r^2)*cos(theta)
here , dA=differential area of ring.
this requires a fairly simple integral of sin cube theta.the simplification is simple .the crucial point was to choose the differential element.
you cannot choose a ring element you will have to choose a disk as your differential element.
DeleteThe radius vector of any point on the ring element makes same angle with a vector, thus giving same dot product value.
DeleteLet vector A be along the x-axis. After that, instead of taking the azimuthal and inclination angle, take a differential ring, it makes the problem fairly easy
ReplyDeleteCan you show the full solution using this method?
DeleteHow about gauss law... Is there any way to use it
ReplyDeleteIn each quarter of this 3d system charge is varying so resultant.direction of.electric field in which direction,,,,,can you show????
ReplyDelete