Question

Question 2.20: 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.

Class 12 - Physics - Electrostatic Potential And Capacitance Electrostatic Potential And Capacitance Page-89

Answer 1

as the spheres 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, $V_a=V_b$
we know, potential , $P=\frac{1}{4\pi\epsilon_0}\frac{q}{r}$

so, $V_a=\frac{1}{4\pi\epsilon_0}\frac{q_a}{a}$
and $V_b=\frac{1}{4\pi\epsilon_0}\frac{q_b}{b}$
$\because\:\frac{q_a}{a}=\frac{q_b}{b}$

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, Ea > Eb

e.g., 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.

Answer 2

potential at sphere of radius a = potential at sphere of radius b

or,

we know, potential ,

so,

and

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

now, if b > a then, Ea > Eb

e.g., 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