# 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.

Since both spheres are connected through the wire, their potential will be the same

Let electric field at A and B be $E_A and E_B$.

Now,

$\frac{E_A}{E_B}=\frac{Q_A}{Q_B}*\frac{b^2}{a^2}$

also

$\frac{Q_A}{Q_B}=\frac{C_aV}{C_BV}$

Also

$\frac{C_A}{C_B}=\frac{a}{b}$

Therefore,

$\frac{E_A}{E_B}=\frac{ab^2}{ba^2}=\frac{b}{a}$

Therefore ratio of the electric field is b/a.

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