The electric field in a region is radially outward with magnitude E = Ar. A sphere of radius 20 cm, centered at the origin, would be containing an amount of charge_____.
(Take A = 100 Vm−2)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8.89 × 10−11 C
By Gauss’s law, the flux through a closed surface equals the enclosed charge divided by \( \varepsilon_0 \). The field is radial and has the same magnitude \( E = Ar \) at every point of the sphere, so the flux is simply E times the area of the sphere.
\( \displaystyle \phi = E \times 4\pi r^{2} = Ar \times 4\pi r^{2} = 4\pi A r^{3} \)
\( \displaystyle q = \varepsilon_0 \phi = 4\pi\varepsilon_0 A r^{3} \)
\( r = 20\,\mathrm{cm} = 0.2\,\mathrm{m} \)
\( r^{3} = 8\times10^{-3}\,\mathrm{m^{3}} \)
Using \( 4\pi\varepsilon_0 = \frac{1}{9\times10^{9}} \):
\( \displaystyle q = \frac{100\,\mathrm{V\,m^{-2}} \times 8\times10^{-3}\,\mathrm{m^{3}}}{9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}}} \)
\( \displaystyle q = \frac{0.8}{9\times10^{9}}\,\mathrm{C} \approx 8.89\times10^{-11}\,\mathrm{C} \)