91Ó°ÊÓ

Write the expression for the electric field due to a point charge is,

$\mathbf{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\stackrel{\mathbf{Áåœ}}{\mathbf{r}}$

Here,$q$ is the point charge,${\mathrm{Î\mu }}_0$ is the permittivity of the free space, and$r$ is the distance from the point charge to the field point.

(a)

Write the expression for the surface area of the cavity of radius is,

$A=4{\mathrm{Ï€²¹}}^2$

Write the formula for the surface charge density on the surface of the cavity of radius is,

${\mathrm{σ}}_a=-\frac{q_a}{A}$

Here,${\mathrm{σ}}_a$is the surface charge density of radius a,A is the surface area and$q_a$is the surface charge of the cavity of radius .

Substitute$4{\mathrm{Ï€²¹}}^2$for$A$in above equation.

${\mathrm{σ}}_a=-\frac{q_a}{4{\mathrm{Ï€²¹}}^2}$

Hence, the surface charge density on the surface of the cavity of radius .

Now, write the expression for the surface area of the cavity of radius$b$is,

$A=4{\mathrm{Ï€²ú}}^2$

Write the formula for the surface charge density on the surface of the cavity of radius$b$is,

${\mathrm{σ}}_b=-\frac{q_b}{A}$

Here, ${\mathrm{σ}}_b$is the surface charge density of radius $b$, $A$is the surface area and $q_b$ is the surface charge of the cavity of radius b .

Substitute $4{\mathrm{Ï€²ú}}^2$for A in above equation.

${\mathrm{σ}}_b=-\frac{q_b}{4{\mathrm{Ï€²ú}}^2}$

Hence, the surface charge density on the surface of the cavity of radius b is.

${\mathrm{σ}}_b=-\frac{q_b}{4{\mathrm{Ï€²ú}}^2}$

The total charge generated on the cavity as a result of electrostatic induction is equal to and opposite to the charge inside the cavity.

$q_{induced}=-q_a=-q_b$

Total charge appears on the surface of the conducting sphere due to induction. It is given by,

$q_{induced}=q_a+q_b$

Write the expression for the surface area of the conducting sphere of radius$R$is,

$A=4{\mathrm{Ï€¸é}}^2$

Now, write the formula for the surface charge density on the surface of the conducting sphere is,

${\mathrm{σ}}_R=\frac{q_a+q_b}{A}$

Substitute$4{\mathrm{Ï€²ú}}^2$ for A in above equation.

role="math" localid="1657955172242" ${\mathrm{σ}}_R=\frac{q_a+q_b}{4{\mathrm{Ï€¸é}}^2}$

Thus, the surface charge density on the surface of the conducting sphere is${\mathrm{σ}}_R=\frac{q_a+q_b}{4{\mathrm{Ï€¸é}}^2}$ .

b)

Write the expression for the total charge appears on the surface of the conducting sphere due to induction. It is given by,

$q_{induced}=q_a+q_b$

Thus, the formula for the electric field outside the conductor is,

$E=\frac{q_a+q_b}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^{\mathrm{r}}}\stackrel{Áåœ}{r}$

Here, E is the electric field and${\mathrm{Î\mu }}_0$ is the permittivity for free space.

c)

Let’s consider that $q_a$ is the charge within the cavity of radius a. Thus, the electric field in cavity of radius a is,

$E_a=\frac{q_a}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}{\stackrel{Áåœ}{r}}_a$

Let’s consider that $q_b$is the charge within the cavity of radius $b$. Thus, the electric field in cavity of radius$b$ is,

$E_b=\frac{q_b}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\stackrel{Áåœ}{r_b}$

Here,$\stackrel{Áåœ}{r_a}$,$\stackrel{Áåœ}{r_b}$are the unit vectors from center of cavity of a,brespectively.

Thus, the electric fields with in each cavity are role="math" localid="1657955861460" $E_a=\frac{q_a}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\stackrel{Áåœ}{r_a}\mathrm{and}{E}_b=\frac{q_b}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}\stackrel{Áåœ}{r_b}$.

d)

Consider the charges $q_a$and $q_b$both only sense the electric field from $-q_a$and $-q_b$their inner shells, and the electric field from these smeared charges cancels out because it pulls with the same force in all directions, so the charges feel no force in any direction.

Hence, the force on$q_a$ and$q_b$ are zero.

e)

Part (a) shows that the results for ${\mathrm{σ}}_a$and ${\mathrm{σ}}_b$do not change when a third charge ${\mathrm{σ}}_c$is brought closer to the conductor. This is due to the absence of an electric field within the hollow conductor. Because of charge conservation, the result for ${\mathrm{σ}}_R$in section (a) will alter when a third charge${\mathrm{σ}}_c$is brought closer to the conductor.

Because of charge conservation, the result for ${\mathrm{σ}}_R$will change when a third charge ${\mathrm{σ}}_c$is brought closer to the conductor. Because the electric field in portion (b) is affected by the surface charge density ${\mathrm{σ}}_R$of the conducting sphere. As a result, when the third charge ${\mathrm{σ}}_C$is brought closer to the conductor. Then the result for the electric field outside the conductor in section (b) changes.

When a third charge ${\mathrm{σ}}_c$is brought closer to the conductor, the results for $E_a$and $E_b$do not change in part (c). This is due to the fact that the charges $q_a$and $q_b$do not change.

When a third charge$q_c$ is brought closer to the conductor, the result for force on each charge does not change in section (d). This is due to the charges experiencing no force in either direction.