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Prove that the field on the surface of a sphere carrying a uniformly distributed electric charge is equal to that which would have been established, if the entire charge were concentrated in the centre of the sphere.

Answer» <html><body><p></p>Solution :Construct a second sphere <a href="https://interviewquestions.tuteehub.com/tag/around-5602275" style="font-weight:bold;" target="_blank" title="Click to know more about AROUND">AROUND</a> the sphere under consideration and suppose it carries a charge <a href="https://interviewquestions.tuteehub.com/tag/equal-446400" style="font-weight:bold;" target="_blank" title="Click to know more about EQUAL">EQUAL</a> in magnitude, but opposite in sign (Fig. 24.14). According to the result <a href="https://interviewquestions.tuteehub.com/tag/obtained-7273275" style="font-weight:bold;" target="_blank" title="Click to know more about OBTAINED">OBTAINED</a> in the <a href="https://interviewquestions.tuteehub.com/tag/previous-592857" style="font-weight:bold;" target="_blank" title="Click to know more about PREVIOUS">PREVIOUS</a> problem, the charge of the outer sphere does not create a field inside it. Therefore the field between the spheres is created only by the charge of the internal sphere. If there is <a href="https://interviewquestions.tuteehub.com/tag/little-1075899" style="font-weight:bold;" target="_blank" title="Click to know more about LITTLE">LITTLE</a> difference between the radii of these spheres (i.c. if `R_(1)-R lt lt R)` , the field in between will be almost homogeneous, and its strength will be (see 37.5) <br/> `E=(sigma)/(epsi_(0))=(Q)/(epsi_(0)S)=(Q)/(4pi epsi_(0)R^(2))`</body></html>


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