Question

Two charged conducting spheres of radii a and b connected by a wire. What is the ratio of the electric field at the surface of the two spheres? Use the result to explain why the charge density on the sharp and pointed ends of a conductor is higher than on its fatter portions.

Answer 1

Calculate the electric field at the surface of the two spheres and then take the ratio of the electric field in terms of charge and radius of the sphere. As the spheres are connected through the wire their potential is the same on the surface. Assuming some capacitance of each sphere calculates the ratio of charges in terms of the radii of the spheres to get the electric field ratio.

Complete step by step answer:

Let the radius of sphere A as $R_A$ and the capacitance of the sphere as $C_A$ and let the radius of sphere b as $R_B$ and the capacitance of the sphere as $C_B$.
Now the electric field at the surface of the sphere is given by
$E = \dfrac{Q}{4\pi \epsilon_o r^2}$
Where Q is the charge and the radius of the sphere is r and E is the electric field
Now the ratio of the electric fields can be calculated as :
$\dfrac{E_A}{E_B} = \dfrac{Q_A}{4\pi \epsilon_o R_A ^2} \times \dfrac{4\pi \epsilon_o R_B ^2}{Q_B}$
We get : $\dfrac{E_A}{E_B} = \dfrac{Q_A}{Q_B} \times \dfrac{b^2}{a^2}$
Now we have calculated the ratio of the electric field in terms of charge and radius.
We know for the capacitor the relation between capacitance and voltage is given as :
$Q = CV$ where C is capacitance and V is the voltage Q is the charge on the capacitor.
Since the voltage is the same we get the ratio of the charges on the two spheres as
$\dfrac{Q_A}{Q_B} = \dfrac{C_A V }{C_B V} = \dfrac{R_A}{R_B}$
Know we found the ratio of the capacitance in the ratio of the charges of the two spheres in terms of the radius
substituting the ratio of the two charges in terms of the radius in the ratio the electric fields we get
$\dfrac{E_A}{E_B} = \dfrac{a}{b} \dfrac{b^2}{a^2} = \dfrac{b}{a}$
So the ratio of the electric field on the two surfaces $\dfrac{R_B}{R_A}= \dfrac{b}{a}$

Now you can see the ratio of the electric field of the sphere A and sphere B is obtained as the ratio of the radii of the sphere B and sphere A.
So we got the electric field at the surface of the spear is inversely proportional to the radius of the sphere.
So $E \propto \dfrac{1}{R}$ where R is the radius of the sphere
Now we can assume a sharp and pointed end as a sphere of very small radius and a flat portion as a sphere of much larger radius and hence the electric field at the surface of the sphere is inversely proportional to the radius of the sphere we get the electric field of the sharp-pointed edge as very large compared to the electric field of the flat portion.

Note:
When two spheres are connected by wire we should be able to conclude that the potential difference between the two spheres is zero if the potential of the two squares is the same with respect to the ground otherwise we won't be able to proceed further. noting down this point will help you solve this problem quickly.