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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? - Physics

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.

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Solution

let a be the radius of a sphere A, QA be the charge on the sphere, and CA be the capacitance of the sphere. Let b be the radius of a sphere B, QB be the charge on the sphere, and CB 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 EB be the electric field of sphere B. Therefore, their ratio,

`E_A/E_B=Q_A/(4piin_0xxa_2)xx(b^2xx4piin_0)/Q_B`

`E_A/E_B=Q_A/(a_2)xx(b^2)/Q_B`

However, `Q_A/Q_B=(C_AV)/(C_BV)`

Putting the value of (2) in (1), we obtain

`E_A/E_B=(ab^2)/(ba^2)=b/a`

Therefore, the ratio of electric fields at the surface is`b/a`.

A sharp and pointed end can be treated as a sphere of very small radius and a flat portion behaves as a sphere of much larger radius.Therefore, charge density on sharp and pointed ends of the conductor is much higher than on its flatter portions.

Concept: Potential Energy in an External Field - Potential Energy of a System of Two Charges in an External Field
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APPEARS IN

NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Q 20 | Page 90
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