91影视

The given data can be listed below as:

• The radius of the solid plastic sphere is,$R_1$
• The charge of the solid plastic sphere is,$-Q_1$$-Q_1$
• The inner radius of the concentric spherical metal shell is,$R_2$
• The inner surface charge of the concentric spherical metal shell is,$Q_2$
• The outer radius of the concentric spherical metal shell is,$R_3$
• The outer surface charge of the concentric spherical metal shell is,$Q_3$

The electric flux is described as the product of the electric field and the special surface鈥檚 area. Moreover, the electric flux is also referred to as the division of the charge of an object with the permittivity.

The gaussian surface at the point $r<R_1$will take no charge. However, by the symmetry of the Gaussian surface, the electric field is equal to zero at all the points inside the plastic sphere.

Thus, the electric field inside the plastic sphere is 0.

A Gaussian surface as a sphere has been considered at the radius $R_1<r<R_2$. However, the magnitude of the electric field is the same at all the points on the surface, and also it is directed towards the center.

The equation of the electric flux is expressed as:

$蠒=EA$

鈥�(颈)

Here, $E$is the electric field and $A$is the spherical surface鈥檚 area.

For the charge $-Q_1$, the equation of the electric flux is expressed as:

$蠒=\frac{-Q_1}{{蔚}_0}$

鈥�(颈颈)

Here, is the charge of the solid plastic sphere and is the permittivity.

Equating equation (i) and (ii),

$EA=\frac{-Q_1}{{蔚}_0}$

Substitute for in the above equation.

role="math" localid="1656933780132" $$\begin{gathered} E.4蟺r^2=\frac{-Q_1}{{蔚}_0} \\ E=\frac{-Q_1}{{蔚}_{0}4蟺r^2} \end{gathered}$$

Thus, the electric field in the air gap and its direction is negative.

As inside the metal, the value of the inner surface charge of the concentric spherical metal shell and charge of the solid plastic sphere is equal, that is $Q_1=Q_2$. Hence, at this point $R_2<r<R_3$, the charge inside any type of spherical surface is zero.

Thus, the electric field in the metal is $0$.

For $r>R_3$,

For the charge $Q_1$, $Q_2$and $Q_3$, the equation of the electric flux is expressed as:

$蠒={\frac{-Q_1+Q_2+Q_3}{{蔚}_0}}^{}$

鈥�(颈颈颈)

Here, is the charge of the solid plastic sphere, the inner and the outer surface charge of the concentric spherical metal shell are $Q_2$and $Q_3$and ${蔚}_0$is the permittivity.

Equating the equation (i) and (iii),

$EA=\frac{-Q_1+Q_2+Q_3}{{蔚}_0}$

Substitute $4蟺r^2$for $A$and $Q_1$for $Q_2$in the above equation.

$E路4蟺r^2=\frac{-Q_1+Q_1+Q_3}{{蔚}_0}$

$E路4蟺r^2=\frac{Q_3}{{蔚}_0}$

$E=\frac{Q_3}{{蔚}_{0}4蟺r^2}$

Thus, the electric field outside the metal is $\frac{Q_3}{{蔚}_{0}4蟺r^2}$and it is directed outwards from the sphere鈥檚 center.

Here the value of the charge $-Q_1$is given as $-5\mathrm{nC}$. As the electric field inside the particular conductor is zero, then for the radius $R_2<r<R_3$, the net charge inside the sphere is zero. Hence, the value of the second charge becomes $Q_1=Q_2$.

Thus, the value of the charge $Q_2$is $5\mathrm{nC}$.

The molecular polarization is described as the ratio of the dipole moment to the local electric field. However, inside a particular crystalline solid, the dipole moment is being considered per unit cell.

Thus, the plastic鈥檚 dipole鈥檚 negative part will be outward and the positive part will be towards the center.