Question

The electric field in a region is radially outward and at a point is given by E = 250 r V/m (where r is the distance of the point from origin). Calculate the charge contained in a sphere of radius 20 cm centred at the origin
(A) $2.22\times {{10}^{-6}}C$
(B) $2.22\times {{10}^{-8}}C$
(C) $2.22\times {{10}^{-10}}C$
(D) Zero

Answer 1

Hint:Electric field has its origin due to the presence of the electric charge. A static charge can produce a field around it. The area of influence of the field depends upon the magnitude of the charge. Also, a charge in motion produces electric fields. Not only a single charge, but a distribution of charge, say surface charge can produce electric fields around it. In this problem we can make use of Gauss law to find the charge.

Complete step by step answer:
Electric field, $E=250rV/m$where r is the distance from centre. So, at the surface of the sphere, $r=20cm=0.2m$. Thus, Electric field from any point on sphere is given by $250\times 0.2=50V/m$
Radius of the sphere, $r=20cm=0.2m$
Surface area of the sphere, $A=4\pi {{r}^{2}}=4\times 3.14\times {{(0.2)}^{2}}=0.5024{{m}^{2}}$
Gauss law states that the total flux through a closed surface is \[\dfrac{1}{{{\varepsilon }_{0}}}\]times the total charge inside that surface.
$\Rightarrow EA=\dfrac{q}{{{\varepsilon }_{0}}}$
$\Rightarrow 50\times 0.5024=\dfrac{q}{{{\varepsilon }_{0}}}$
We know the value of permittivity of free space is $8.85\times {{10}^{-12}}$
$\Rightarrow q=50\times 0.5024\times 8.85\times {{10}^{-12}}$
$\therefore q=2.22\times {{10}^{-10}}C$

So, the correct option is C.

Note:Had it been a case like a charge was situated inside the cube then we would have to apply gauss law to find the net flux through the cube. Gauss law states that the total flux through a closed surface is \[\dfrac{1}{\varepsilon }\]times the total charge inside that surface. Flux is given by the dot product of electric field vector and the area vector, since it involves dot product it is a scalar quantity.