91Ó°ÊÓ

The given data can be listed below as:

• The radius of the thin plastic spherical shell is $5\mathrm{cm}×\left(\frac{{10}^{-2}}{1\mathrm{cm}}\right)=0.05\mathrm{cm}$.
• The charge on the spherical shell is ${\mathrm{q}}_1=-25\mathrm{nC}×\left(\frac{{10}^{-9}}{1\mathrm{nC}}\right)=-25×{10}^{-9}\mathrm{C}$.
• The radius of the concentric thin plastic spherical shell is $8\mathrm{cm}×\left(\frac{{10}^{-2}}{1\mathrm{cm}}\right)=0.08\mathrm{cm}$.
• The charge on the concentric thin plastic spherical shell is ${\mathrm{q}}_2=+64\mathrm{nC}×\left(\frac{{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)=64×{10}^{-9}\mathrm{C}$.

The magnitude of the electric field is directly proportional to the charge and inversely proportional to the square of their distances.

As the radius of the shells are and respectively, then for a distance of 3 cm, the point lies inside both the shells.

Thus, the electric field at a distance of is .

As the distance 7 cm lies inside the concentric thin plastic spherical shell, then the electric field of the sphere in that distance is 0. For the thin plastic spherical shell, the point lies outside, hence this sphere will get an inward electric field.

The equation of the magnitude of the electric field is expressed as:

$E=k\frac{q_1}{r^2}$

Here, k is the electric field constant that has the value of $9×{10}^{9}N.m^2/C^2$, is the charge of the thin plastic spherical shell and r is the distance of the electric field from the center.

Substitute the values in the above equation.

role="math" localid="1656935440585" $$\begin{gathered} \mathrm{E}=9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×\frac{-25×{10}^{-9}\mathrm{C}}{{\left(7\mathrm{cm}\left({\displaystyle \frac{10\mathrm{m}}{100\mathrm{cm}}}\right)\right)}^2} \\ =9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×\frac{-25×{10}^{-9}\mathrm{C}}{4.9×{10}^{-3}{\mathrm{m}}^2} \\ =9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×-5.102×{10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \\ =-4.59×{10}^4\mathrm{N}/\mathrm{C} \end{gathered}$$

The equation of the magnitude of the net electric field is expressed as:

$E=k\frac{q_1}{r^2}+k\frac{q_2}{r^2}$

Here, $q_1$is the charge of the thin plastic spherical shell and $q_2$is the charge of the concentric thin plastic spherical shell, k is the electric field constant that has the value of $9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2,q$, is the charge of the electric field and is the distance of the electric field from the center.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{E}=9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×\frac{-25×{10}^{-9}\mathrm{C}}{{\left(10\mathrm{cm}\left({\displaystyle \frac{10\mathrm{m}}{100\mathrm{cm}}}\right)\right)}^2} \\ =9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×\frac{64×{10}^{-9}\mathrm{C}}{{\left(10\mathrm{cm}\left(\frac{10\mathrm{m}}{100\mathrm{cm}}\right)\right)}^2} \\ =9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×\left(\frac{-25×{10}^{-9}\mathrm{C}}{0.01{\mathrm{m}}^2}+\frac{64×{10}^{-9}\mathrm{C}}{0.01{\mathrm{m}}^2}\right) \\ =-9×{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2×3.9×{10}^{-6}\mathrm{C}/{\mathrm{m}}^2 \\ =3.51×{10}^4\mathrm{N}/\mathrm{C} \end{gathered}$$

The positive sign indicates that the electric field points to the outward portion of the sphere.

Thus, the electric field at a distance of 3 cm, 7 cm and 10 cm from the center are 0, $-4.59×{10}^4\mathrm{N}/\mathrm{C}$which is radially inward and $3.51×{10}^4\mathrm{N}/\mathrm{C}$which is radially outward.