Question

if one penetrates a uniformly charged solid sphere, the electric field E:
A. Increases
B. Decreases
C. Is zero at all poles
D. Remains same as at the surface

Answer 1

The solid sphere is uniformly charged. For a uniformly charged sphere, the charge resides throughout the solid surface. Consider a Gaussian surface inside the sphere. Find the Electric field on the inner side of the sphere.

Complete step by step answer:
It is given to us that the solid sphere is uniformly charged. As mentioned in the hint, for a uniformly charged body, the charge is uniformly distributed throughout the sphere.
As one penetrates a uniformly charged solid sphere, he enters inside the solid sphere. Therefore, we need to find the Electric field on the inside of the sphere. Now, let us consider a Gaussian surface inside the solid sphere in the shape of a sphere. We can find the electric field using Gauss law.

Gauss law states describes the static electric field generated by a distribution of electric charges. Gauss law states, the electric flux through any closed surface is proportional to the total electric charge enclosed by this surface.
Mathematically, we can write this as:
$E \times A = \dfrac{{{q_{enclosed}}}}{{{\varepsilon _0}}}$
Where $E$ is the required electric field
$A$ is the surface area of the gaussian surface
${q_{enclosed}}$ denotes the charge enclosed
${\varepsilon _0}$ is a constant.
Let $R$ be the radius of the given solid sphere.

Now, the area of the sphere is $A = 4\pi {r^2}$ where $r$ is the radius of the gaussian sphere; such that $r < R$ as we are inside the sphere.
$\Rightarrow E \times (4\pi {r^2}) = \dfrac{{{q_{enclosed}}}}{{{\varepsilon _0}}}$
Now, for $r < R$ the gaussian surface will enclose less than the total charge and hence the electric field will also be less. The electric field in such cases is given by:
$\therefore E = \dfrac{{Qr}}{{4\pi {\varepsilon _0}{R^3}}}$
Where $Q$ is the total charge.
As it is evident from the formula, that for $r < R$ the electric field will decrease as we move inside the sphere.

Therefore, option B is the correct option.

Note: Do remember that electric field for a uniformly charged solid sphere inside the sphere is given as $E = \dfrac{{Qr}}{{4\pi {\varepsilon _0}{R^3}}}$
And the electric field for any point outside the sphere such that $r > R$ is given as $E = \dfrac{Q}{{4\pi {\varepsilon _0}{R^2}}}$ . For points outside the sphere the electric field will decrease as we move away from the sphere.