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 surface of 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 fatter portions ?

Explanation

As both the charged spheres are connected through a wire, the electric potential on the both the spheres will become same after certain time.
Let the capacitance of the sphere with radius a be `C_(1) ` and that with radius b be `C_(2)`.
Thus,
Charge on the sphere with radius a, `Q_(1)=C_(1)V` and
Charge on the sphere with radius b, `Q_(2)=C_(2)V`
As we know: Capacitance of sphere` prop` Radius of sphere
`rArr Q_(1)prop a and Q_(2) prop b`
Electric field at the surface of both the spheres can be given as:
`(E_(1))/(E_(2))=(sigma_(1))/(sigma_(2))=(((Q_(1))/(4pi epsilon_(0)a^(2))))/(((Q_(2))/(4pi epsilon_(0)b^(2))))=(Q_(1)b^(2))/(Q_(2)a^(2))`
As surface charge density on a sphere with larger radius is less and the sphere with a smaller radius will have more charge density. Similarly, a sharp pointed object can be considered as s sphere with small radius and the flattened portions as the parts of the sphere with larger radius. Thus, a sharpened or pointed object of the conductor will have more charge density than on its flattened portions.