## Saturday, February 6, 2010

### Irodov Problem 3.16

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,

Subscribe to:
Post Comments (Atom)

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.

Let 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

ReplyDeleteHow about gauss law... Is there any way to use it

ReplyDelete