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Question 20: NCERT Solutions for 12th Class Physics: Chapter 2-Electrostatic Potential and Capacitance
TWO charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Solution:
As the two spheres are connected to each other by wire, so they have same electric potential i.e.,Va = Vb
if b > a, then Ea > Eb
i.e. sphere with smaller radius produces more electric field on its surface. Hence, the charge density on the sharp and pointed ends of conductor is higher than on its flatter portions.
Check the complete chapter with solutions.
NCERT Solutions for 12th Class Physics: Chapter 2-Electrostatic Potential and Capacitance