Question

1. For a thin spherical shell of uniform surface charge density sigma. The magnitude of E at a distance r. when r> R (radius of shell) is


a. E = 4TRO
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b. E = 4Ro
Ro

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d. E= 41 Ro2
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Answer 1

Given :

The uniform surface charge density of the given thin spherical shell = $\sigma$

Radius of this shell = R

To Find ;

The magnitude of electric field at a distance r ( r > R ) from the centre of the shell = ?

Solution :

The surface area of this spherical shell will be = $4 \pi R^2$

So, the total charge on the shell = $\sigma .4 \pi R^2$

By Gauss law we have :

$\int \vec E . \vec{ds} = \frac{Q}{\epsilon_0}$

So, $E \times 4 \pi r^2 = \frac{\sigma \times 4 \pi R^2}{\epsilon_0}$

Or, $E = \frac{\sigma .R^2}{\epsilon_0.r^2}$

Hence , E at a distance r from the centre of the shell where r > R is $E = \frac{\sigma .R^2}{\epsilon_0.r^2}$