Question

The electric field at point P just outside the outer surface of a hollow spherical conductor of inner radius 10 cm and outer radius 20 cm has magnitude 459 N/C and is directed outward. When an unknown point charge Q is introduced into the center of the sphere, the electric field at P is still directed outward but is now 181 N/C. (a) What is the net charge on and enclosed by the outer spherical surface before Q is introduced?

Answer 1

Answer:

$2.042\times 10^{-9}\ C.$

Explanation:

Given:

• Electric field just outside the given sphere, E = 459 N/C.
• Inner radius of the shell, $\rm r_i$ = 10 cm = 0.1 m.
• Outer radius of the shell, $\rm r_o$ = 20 cm = 0.2 m.
• Electric field outside the shell after introducing the charge Q, $\rm E_f$ = 181 N/C.

Before introducing the charge Q,

Consider a Gaussian sphere of same radius as the outer radius of the shell, concentric with the given shell,

Applying Gauss' law over this surface,

$\oint \vec E \cdot d\vec A = \rm \dfrac{q}{\epsilon_o}.$

where,

• $\epsilon_o$ is the electrical permittivity of free space having value $9\times 10^9\rm \ Nm^2/C^2$.

• $d\vec A$ is the surface area element of the sphere, directed along the normal to the plane of the surface.
• $\rm q$ is the net charge enclosed by the Gaussian surface.

The directions of the area element and the electric field, both are directed outward the surface, therefore, $\vec E \cdot d\vec A = E\ dA$.

The LHS of the equation of the Gauss' law is given as

$\oint \vec E\cdot d\vec A=\oint E\ dA =E\oint dA$

$\oint dA$ is the surface area of the Gaussian sphere = $\rm 4\pi r_f^2$

Therefore, using the equation of the Gauss' law,

$\oint \vec E\cdot d\vec A=\rm E\ 4\pi r_f^2=\dfrac{q}{\epsilon_o}\\\Rightarrow q =E\ 4\pi r_f^2\times \epsilon_o\\=459\times 4\pi \times (0.2)^2\times 8.85\times 10^{-12}\\=2.042\times 10^{-9}\ C.$

It is the net charge enclosed by the outer surface before Q was introduced.