Question

Figure 23-27 shows four solid spheres, each with charge $\mathit{Q}$uniformly distributed through its volume. (a) Rank the spheres according to their volume charge density, greatest first. The figure also shows a point for each sphere, all at the same distance from the center of the sphere. (b) Rank the spheres according to the magnitude of the electric field they produce at point $\mathit{P}$, greatest first.

Short answer

1. The rank of the spheres according to their volume charge densities is.$a>b>c>d$
2. The rank of the spheres according to the magnitude of the electric field at point P is.$E_a>E_b>E_c>E_d$

Step-by-step solution

The given data:

Figure 23-27 showing four solid spheres, each with charge Q uniformly distributed through its volume are given.

Understanding the concept of volume charge density and electric field:

The volume charge density of a body is given by the amount of charge carries by the body divided by its volume. Now, using this volume charge density relation with the electric field of the sphere, the rank of the spheres can be determined for different positions of the point P.

Formulae:

The volume charge density of a sphere,

$\rho =\frac{q}{\frac{4}{3}\pi R^3}$….. (i)

The electric field at point inside the sphere,

$E=\frac{\rho r}{3{\epsilon }_0}(r<R)$ ….. (ii)

(a) Calculation of the rank of spheres according to volume charge densities:

From equation (i), it can see that the volume charge density of a sphere$\left(\rho \right)$ is inversely proportional to the radius of the sphere$\left(R\right)$. That is given as:

$\rho \propto \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$R^3$}\right.$

Thus, as the radius of the sphere increases the volume charge density decreases.

The rank of the radius of the spheres from the figure can be given as:

$d>c>b>a$

Hence, the rank of the spheres according to charge densities is$a>b>c>d$.

(b) Calculation of the rank of the spheres according to the magnitude of electric field:

For situation a and b, the points location is outside the sphere.

Thus, their electric field outside the sphere is same as that at the surface of the sphere can be given using equation (ii) for the same charge spread:

$E=\frac{q}{4\pi {\epsilon }_0{r}^2}$ …… (iii)

Here, r is the distance of point P from the center of the sphere,

Again, the radius of the sphere is small than that of sphere b, thus the electric field value can be given as:

$E_a>E_b$

The sphere a and b has the same magnitude of the electric field, because for both the sphere, the charge q and distance r are the same.

Now, for the spheres and , the points are located inside the sphere, thus, the electric field is given by equation (ii).

Therefore,

$E\propto \rho r$

If you assume that the point to be equally at distance from the center, then the electric field would be given by:

$E\propto \raisebox{1ex}{$r$}\!\left/ \!\raisebox{-1ex}{$R^3$}\right.$ ….. (iv)

Thus, the rank of electric fields for c and d would be

Because the sphere is having the volume charge density greater than the sphere d .

Now, considering equations (iii) and (iv), the final rank of the spheres according to fields is given as:

$E_a>E_b>E_c>E_d$

Hence, the require rank is.$E_a>E_b>E_c>E_d$