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. $ 4 \,\pi\varepsilon_0\, Aa^3 $
• B. $ \varepsilon_0\, Aa^3 $
• C. $ 4 \, \pi\varepsilon_0 \, Aa^2 $
• D. $ A \, \varepsilon_0 \,a^2 $

Answer 1

The Correct Option is A

To solve this problem, we need to determine the charge contained within a sphere of radius \(a\) when the electric field is given by \(E = Ar\), where \(A\) is a constant.

Step-by-Step Solution 

1. The electric field \(E\) is acting radially outward. In such cases, Gauss's Law is applicable, which states:\)

\(\Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\)

• Here, \(\Phi\) is the electric flux through a closed surface, \(Q_{\text{enc}}\) is the charge enclosed, and \(\varepsilon_0\) is the permittivity of free space.
• Because the electric field is uniform over the spherical surface at a distance \(r\), the flux through the surface is:

\(\Phi = E \cdot 4\pi r^2\)

1. Substitute the given electric field, \(E = Ar\), into the expression for flux:

\(\Phi = Ar \cdot 4\pi r^2 = 4\pi A r^3\)

1. Apply Gauss's Law:

\(\frac{Q_{\text{enc}}}{\varepsilon_0} = 4\pi A r^3\)

1. Solve for the charge \(Q_{\text{enc}}\):

\(Q_{\text{enc}} = 4\pi \varepsilon_0 Ar^3\)

1. Since we are considering a sphere of radius \(a\), substitute \(r = a\):

\(Q_{\text{enc}} = 4\pi \varepsilon_0 Aa^3\)

Conclusion

Therefore, the charge contained within the sphere of radius \(a\) is given by \(4\pi \varepsilon_0 Aa^3\).

The correct answer is thus: \(4\pi \varepsilon_0 Aa^3\).