Answer:
The magnitude of Electric Field is [tex]E=\dfrac{Qr}{4\pi \epsilon_0 R^3}[/tex]
Explanation:
Given:
Let consider a Gaussian surface at a distance of r such that 0<r>R in the shape of sphere such that the electric Field due to this E and it is radially outwards.
The charge inside this Gaussian surface volume we have , [tex]q_{in}=\dfrac{Qr^3}{R^3}[/tex]
Now using Gauss Law we have
[tex]E\times4\pi r^2=\dfrac{q_{in}}{\epsilon_0}\\E\times4\pi r^2=\dfrac{\dfrac{Qr^3}{R^3}}{\epsilon_0}\\E=\dfrac{Qr}{4\pi \epsilon_0 R^3}[/tex]