Question

Figure 21-16 shows three situations involving a charged particle and a uniformly charged spherical shell. The charges are given, and the radii of the shells are indicated. Rank the situations according to the magnitude of the force on the particle due to the presence of the shell, greatest first.

Short answer

The rank of the situations according to the magnitude of the force on the particle due to the presence of the shell is $\left|{\mathrm{F}}_2\right|=\left|{\mathrm{F}}_3\right|>\left|{\mathrm{F}}_1\right|$.

Step-by-step solution

Stating the given data

Figure 21-16 represents three situations involving a charged particle and a uniformly charged spherical shell, with their charges and radii being indicated.

Understanding the concept of force on a particle due to the shell

The net force will be zero inside the conducting shell, as all the charges get distributed over the surface and thus have no effect within the shell body provided that no electric flux passes within the shell. But for the particles that are outside the shell radius, the net force due to the shell is given by the usual electrostatic force value.

Formula:

The magnitude of the electrostatic force due to the two charges is $\left|\mathrm{F}\right|=\frac{\mathrm{k}\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$. (i)

 Step 3: Calculation of the rank according to the magnitude of the force on the particle due to the shell

From the given figure, the charges of the particles in the three situations are $+6\mathrm{q},+2\mathrm{q},-\mathrm{q}$while the charges on the shells are $+5\mathrm{Q},-4\mathrm{Q},+8\mathrm{Q}$respectively.

For situation (a):

The charge particle is at distance $\mathrm{d}<2\mathrm{R}$.

Thus, the electric field inside the shell will be zero. So, from equation (ii), the electrostatic force on the particle will be zero, that is: $\left|{\mathrm{F}}_1\right|=0$

For situation (b):

The charge particle is at distance $\mathrm{d}>\mathrm{R}$.

Thus, the magnitude of the electrostatic force at the charge point $+2\mathrm{q}$due to the shell is obtained using equation (i) as follows:

$\begin{array}{c}\left|{\mathrm{F}}_2\right|=\frac{\mathrm{k}|-4\mathrm{Q}||+2\mathrm{q}|}{{\mathrm{d}}^2}\\ =\frac{8\mathrm{kQq}}{{\mathrm{d}}^2}\end{array}$

For situation (c):

The charge particle is at distance $\mathrm{d}>\mathrm{R}/2$.

Thus, the magnitude of the electrostatic force at the charge point$-q$due to the shell will be given using equation (i) as follows:

$\begin{array}{c}\left|{\mathrm{F}}_3\right|=\frac{\mathrm{k}|+8\mathrm{Q}||-\mathrm{q}|}{{\mathrm{d}}^2}\\ =\frac{8\mathrm{kQq}}{{\mathrm{d}}^2}\end{array}$

Hence the rank of the situations according to the magnitude of forces is $\left|{\mathrm{F}}_2\right|=\left|{\mathrm{F}}_3\right|>\left|{\mathrm{F}}_1\right|$.