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 - Physics

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Numerical

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 EA be 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_0 xx "a"_2) xx ("b"^2 xx 4piin_0)/"Q"_"B"`

`"E"_"A"/"E"_"B" = "Q"_"A"/"Q"_"B" xx ("b"^2)/"a"^2` ......(1)

However, `"Q"_"A"/"Q"_"B" = ("C"_"A""V")/("C"_"B""V")`

And, `("C"_"A")/("C"_"B") = "a"/"b"`

∴ `"Q"_"A"/"Q"_"B" = "a"/"b"` .......(2)

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

∴ `"E"_"A"/"E"_"B" = "a"/"b""b"^2/"a"^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 a very small radius and a flat portion behaves as a sphere of a much larger radius. Therefore, the 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 Physics Part 1 and 2 Class 12
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 2.20 | Page 88
NCERT Class 12 Physics Textbook
Chapter 2 Electrostatic Potential and Capacitance
Exercise | Q 20 | Page 90
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