Question

a point charge is placed at the centre of hollow conducting sphere initial radius R and Outer radius 2r the ratio of the surface charge density of inner surface so that outer surface will be

Answer 1

Answer: -4:1

Explanation: Redistribution of charges.

Answer 2

The ratio of the surface charge density of inner surface and outer surface is -4:1.

Explanation:

Given that,

Inner radius = R

Outer radius = 2R

We need to calculate the surface charge density of inner surface

Using formula of charge density

$\sigma_{in}=\dfrac{-q}{4\pi r^2}$

Put the value into the formula

$\sigma_{in}=\dfrac{-q}{4\pi\times R^2}$....(I)

We need to calculate the surface charge density of outer surface

Using formula of charge density

$\sigma_{out}=\dfrac{+q}{4\pi r^2}$....(I)

Put the value into the formula

$\sigma_{out}=\dfrac{+q}{4\pi\times(2R)^2}$....(II)

Divided equation (I) by equation (II)

$\dfrac{\sigma_{in}}{\sigma_{out}}=\dfrac{\dfrac{-q}{4\pi\times R^2}}{\dfrac{+q}{4\pi\times(2R)^2}}$

$\dfrac{\sigma_{in}}{\sigma_{out}}=\dfrac{-4R^2}{R^2}$

$\dfrac{\sigma_{in}}{\sigma_{out}}=\dfrac{-4}{1}$

Hence, The ratio of the surface charge density of inner surface and outer surface is -4:1.

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Topic : charge density

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