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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 of a conductor is higher than on its flatter portions.

$\begin{array}{1 1} \frac{a}{b} \\\frac{b}{a} \\ a=b \\ b > a \end{array}$

Let a be the radius of a sphere $A, Q_A$ be the charge on the sphere, and $C_A$ be the capacitance of the sphere.
Let b be the radius of a sphere $B, Q_B$ be the charge on the sphere, and $C_B$ be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.
Let $E_A$ be the electric field of sphere A and $E_B4 be the electric field of sphere B. Therefore, their ratio,$\large\frac{E_A}{E_B}= \large\frac{Q_4}{4 \pi \in_0 \times a_2} \times \frac{b^2 \times 4 \pi \in_0}{Q_8}\large\frac{E_A}{E_B} =\frac{Q_A}{Q_B} \times \frac{b^2}{a^2}$------------(1) How ever$\large\frac{Q_A}{Q_B} = \frac{C_1 V}{C_B V}$and$\large\frac{C_A}{C_B} = \frac{a}{b}$----------------(2) Putting the value of (2) in (1), we obtain$\large\frac{E_A}{E_B} =\large\frac{a}{b} \frac{b^2}{a^2}=\frac{a}{b}$Therefore, the ratio of electric fields at the surface is$\large\frac{b}{a}\$