Electric_Field_inside_a_Uniformly_Charged_Spherical_Surface

Electric_Field_inside_a_Uniformly_Charged_Spherical_Surface...

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Unformatted text preview: Electric Field inside a Uniformly Charged Spherical Surface The figure above represents a spherical surface over which charge has been uniformly distributed. Let us calculate the electric field at the point a distance R from the sphere's center. [The figure above was taken from a text in which the gravitational force on mass m due to a spherical shell of mass was being calculated.] Consider the electric field at R. From symmetry, the only surviving component of the field will be that which lies along the line connecting the origin to the point m. Notice, also, that there is nothing special about point m due to the spherical symmetry inherent in the distribution of charge on the spherical surface. In what follows, we will ignore (uninteresting) quantities, i.e., those that only multiply the final result (which will be zero, as we shall show). The contribution to the electric field at R from the ring of charge is: (1) (2! r sin " )rd" cos # x2 where 2! r sin " represents the circumference of the ring and rd! is the ring's width. Thus the numerator represents the surface area of the ring. If we were to multiply this times the surface charge density, we would then have the total charge on this particular ring. The cos ! term represents the component of the electric field from the ring that lies along the R direction. All of the charge on the ring is a distance x from the point at m. x cos ! + r cos" = R or (2) cos ! = R " r cos# x The law of cosines gives: (3) x 2 = R 2 + r 2 ! 2Rr cos" , which can be re-arranged to: R2 + r 2 " x 2 (4) r cos! = . 2R We can differentiate (3): (5) sin ! d! = x dx Rr Now, insert (4) into (2) and (5) into (1). Dropping all constant factors that multiply the result leaves the contribution to the electric field at point m due to the ring of charge: (6) ( R2 ! r 2 +1)dx x2 To find the total field, we need to sum the contributions from all rings which leads to the following integral: (7) r+ R r!R "( R2 ! r 2 +1)dx x2 It is left to the reader to demonstrate that the integral (7) equals zero. ...
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