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 the charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal. Let EA be the electric field of sphere A and EB be the electric field of sphere B. Therefore, their ratio, EBEA=4π∈0×a2QA×QBb2×4π∈0 EBEA=QBQA×a2b2 ...(1) However, QBQA=CBVCAV And CBCA=ba ∴QBQA=ba ...(2) Putting the value of (2) in (1), we obtain ∴EBEA=baa2b2=ab Therefore, the ratio of electric fields at the surface is ab. A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius.Therefore, charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.
