Question

A conducting spherical shell is given charge +20. A point charge +Q is kept inside the shell at point O as shown. Now if charge +Q is shifted from point O to point P then magnitude of electric field at point A outside the shell will

Answer 1

Answer:

she'll be=40 I hope you understand please mark me brilliant

Answer 2

Answer:

Magnitude of electric field outside A will remain same irrrespective of the position of charge inside the spherical shell.

Explanation:

Using Gauss law, $\phi = \frac{q_{enclosed}}{\epsilon_o} = \int \vec E.\vec dS$ ----(1)

here, $q_{enclosed}$ is the charge enclosed inside the gaussian surface.

Let's make a gaussian surface at the place outside the sphere where we need to find the electric field.

charge enclosed will be, $q_{enclosed}=20q+q=21q$

using (1), $\frac{21q}{\epsilon_o} = \int \vec E.\vec dS$

$\implies \frac{21q}{\epsilon_o}=E.(4.\pi.r^2)$

$\implies E = \frac{21q}{4\pi r^2}$

Here, the charge enclosed will remain $21q$ irrespective of the position of charge inside the shell. So, electric field will also remain same irrespective of the position of charge inside the shell.

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