91Ó°ÊÓ

A charge ball lies within a hollow metallic sphere of radius R .

The net charges of the ball and the shell for three situations are given.

Using the electric field concept for a hollow metallic sphere, you can get the required rank of the situations. Due to its properties, the charges of a hollow sphere remain at the surface of the sphere. Thus, the charge inside the shell is zero and so its electric field is also zero. Similarly, the electric field at any point at the surface or above surface is given by the surface charges and the distance of the point from the center of the shell.

Formula:

The electric field at any point outside the hollow metallic sphere,

$E=\frac{kQ}{r^2},râ‰\yen R$….. (i)

Where, Q is the outer charge of the shell, K is the Coulomb’s constant, and r is the distance.

The electric field at any point inside a hollow metallic sphere is

$E_{r<R}=0$

Hence, the rank of the situations according to the electric field at halfway is$1=2=3$ .

Consider a Gaussian surface at a distance $2R$ from the center. Here, the distance of the point is given by

$r=2R$

Thus, the point is outside the shell. So, the magnitude of electric fields for the three situations including its outer charge value is given by,

$E=\frac{k(q_{ball}+q_{shell})}{{\left(2R\right)}^2}$

Here, R is the radius of the spherical shell, $q_{ball}$ is the charge on the ball, and $q_{ball}$ is the charge on the shell.

The magnitude of the electric field for situation 1 when the charge on the ball is $+4q$and on the shell is zero.

$\begin{array}{c}{E}_1=\frac{k(q_{ball}+q_{shell})}{{\left(2R\right)}^2}\\ =\frac{k(+4q+0)}{{\left(2R\right)}^2}\\ =\frac{4qk}{4{\left(R\right)}^2}\\ =\frac{qk}{R^2}\end{array}$

The magnitude of the electric field for situation 2 when the charge on the ball is $−6q$and on the shell is $+10q$.

$\begin{array}{c}{E}_2=\frac{k(−6q+10q)}{{\left(2R\right)}^2}\\ =\frac{4qk}{4{\left(R\right)}^2}\\ =\frac{qk}{R^2}\end{array}$

The magnitude of the electric field for situation 3 when the charge on the ball is $+16q$and on the shell is$−12q$.

$\begin{array}{c}{E}_3=\frac{k(+16q−12q)}{{\left(2R\right)}^2}\\ =\frac{4qk}{4{\left(R\right)}^2}\\ =\frac{qk}{R^2}\end{array}$

Hence, the rank of the situations according to field value is$1=2=3$