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  • 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

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)

best_answer

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.

Posted by

Pankaj Sanodiya

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