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.
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.
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 $B$, $Q$ be the charge on the sphere, and $C$ a be the capacitance of the sphere.
since the two spheres are connected with a wire, their potential $(v)$ will become equal.
Let Eabe the electric field of sphere $A$ and $E_{8}$ be the electric field of sphere $B$. Therefore, their ratio,
$\frac{E_{A}}{E_{B}}=\frac{Q_{A}}{4 \pi \epsilon_{0} \times a_{2}} \times \frac{b^{2} \times 4 \pi \epsilon_{0}}{Q_{B}}$
$\frac{E_{A}}{E_{B}}=\frac{Q_{1}}{Q_{B}} \times \frac{b^{2}}{a^{2}}\ldots(i)$
However, $\frac{Q_{A}}{Q_{B}}=\frac{C_{A} V}{C_{B} V}$
And, $\frac{C_{A}}{C_{B}}=\frac{a}{b}$
$\therefore \frac{Q_{A}}{Q_{B}}=\frac{a}{b}\dots (ii)$
Putting the value of $(ii)$ in $(i)$, we obtain
$\therefore \frac{E_{A}}{E_{B}}=\frac{a}{b} \frac{b^{2}}{a^{2}}=\frac{b}{a}$
Therefore, the ratio of electric fields at the surface is $b/a.$
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