Question

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.

Answer 1

it is given that, two charged spheres of radii a and b are connected to each other by wire, so they have same potential.

e.g., potential at sphere of radius a = potential at sphere of radius b

or, $V_a=V_b$.......(1)

we know, potential, $V=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$

so, $V_a=\frac{1}{4\pi\epsilon_0}\frac{q_a}{a}$.....(2)

and $V_b=\frac{1}{4\pi\epsilon_0}\frac{q_b}{b}$.....(3)

from equations (1) , (2) and (3), we get

$\frac{q_a}{a}=\frac{q_b}{b}$

so, the ratio of electric fields are the surface of the two spheres is

$\frac{E_a}{E_b}=\frac{\frac{1}{4\pi\epsilon_0}\frac{q_a}{a^2}}{\frac{1}{4\pi\epsilon_0}\frac{q_b}{b^2}}$

$=\frac{q_a}{q_b}\times\frac{b^2}{a^2}=\frac{b}{a}$

now, if b > a then, $E_a > E_b$

i.e., sphere with smaller radius produces more electric field intensity on its surface .

hence, the charge density on the sharp and pointed ends of conductor is higher than on its flatter portions.