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.

Answers (1)

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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