Question

The electric field in a certain region is acting radially outward and is given by E = Ar. A charge contained in a sphere of radius= a centred at the origin of the field, will be given by:

• A. ϵ₀Aa³
• B. 4π ϵ₀Aa²
• C. A ϵ₀a²
• D. 4π ϵ₀Aa³

Answer 1

Answer: (D)

Step 1: Given that:

The electric field at a certain region is given by

$E=Ar$

the radius of the sphere placed in the electric field E = $a$

Step 2: Formula used:

According to Gauss' law;

The electric flux linked with a closed surface enclosing a charge q is equal to $\frac{1}{{\epsilon }_0}$ times the charge enclosed by the surface.

Mathematically,

$\oint \overrightarrow{E}.\overrightarrow{dS}=\frac{q}{{\epsilon }_0}$

Step 3: Calculation of the charge enclosed by the sphere:

Since, the sphere is placed in the electric field $E=Ar$ ,

The electric field at the surface of the sphere will be calculated by placing $r=a$ that is the radius of the sphere;

Thus, The electric field at sphere is given as;

$E=A\times a$

Using Gauss' law

$\oint \overrightarrow{E}.\overrightarrow{dS}=\frac{q_{enc}}{{\epsilon }_0}$

$\int EdS=\frac{q_{enc}}{{\epsilon }_0}$

$As,theelectricfieldandthesurfaceareavectorofspherearedirectedinsamedirection.$

$E\int dS=\frac{q_{enc}}{{\epsilon }_0}$

$A\times a.4\pi a^2=\frac{q_{enc}}{{\epsilon }_0}$

$q_{enc}=4\pi {\epsilon }_{0}A{a}^3$

Thus,

Option D) $4\pi {\epsilon }_{0}A{a}^3$ is the correct option.