Answer:
The electric field outside the sphere will be [tex]\dfrac{qr}{4\pi\epsilon_{0}R^3}[/tex].
Explanation:
Given that,
Radius of solid sphere = R
Charge = q
According to figure,
Suppose r is the distance between the point P and center of sphere.
If [tex]\rho[/tex] be the volume charge density,
Then, the charge will be,
[tex]q=\rho\times\dfrac{4}{3}\pi R^3[/tex].....(I)
Consider a Gaussian surface of radius r.
We need to calculate the electric field outside the sphere
Using formula of electric field
[tex]\oint{\vec{E}\cdot \vec{dA}}=\dfrac{Q}{\epsilon_{0}}[/tex]
[tex]E\times4\pi r^2=\dfrac{\rho\dotc \dfrac{4}{3}\pi r^3}{\epsilon_{0}}[/tex]
Put the value from equation (I)
[tex]E\times4\pi r^2=\dfrac{qr^3}{\epsilon_{0}R^3}[/tex]
[tex]E=\dfrac{qr}{4\pi\epsilon_{0}R^3}[/tex]
Hence, The electric field outside the sphere will be [tex]\dfrac{qr}{4\pi\epsilon_{0}R^3}[/tex].
