91Ó°ÊÓ

The given data can be listed below as,

• The radius of innermost sphere is,${\mathrm{p}}_1=2 \mathrm{cm}\left(\frac{1 \mathrm{m}}{100\mathrm{cm}}\right)=2×{10}^{-2} \mathrm{m}.$
• The radius of middle sphere is,${\mathrm{p}}_2=5 \mathrm{cm}\left(\frac{1 \mathrm{m}}{100\mathrm{cm}}\right)=5×{10}^{-2}\mathrm{m}$
• The radius of outermost sphere is,${\mathrm{p}}_3=10 \mathrm{cm}\left(\frac{1 \mathrm{m}}{100\mathrm{cm}}\right)=10×{10}^{-2}\mathrm{m}.$
• The charge on innermost sphere is,${\mathrm{q}}_1=6 \mathrm{nC}=6 \mathrm{nC}\left(\frac{1×{10}^{-9} \mathrm{C}}{1 \mathrm{nC}}\right)=6×{10}^{-9} \mathrm{C}$ .
• The charge on middle sphere is,${\mathrm{q}}_2=-4 \mathrm{nC}=-4 \mathrm{nC}\left(\frac{1×{10}^{-9}\mathrm{C}}{1\mathrm{nC}}\right)=-4×{10}^{-9}\mathrm{C}.$ .
• The charge on outermost sphere is, ${\mathrm{q}}_3=8 \mathrm{nC}=8 \mathrm{nC}\left(\frac{1×{10}^{-9} \mathrm{C}}{1\mathrm{nC}}\right)=8×{10}^{-9} \mathrm{C}.$.
• For part (a) the distance of electric field from the centre is${\mathrm{r}}_1=1\mathrm{cm}\left(\frac{1 \mathrm{m}}{100\mathrm{cm}}\right)=1×{10}^{-2} \mathrm{m}.$.
• For part (b) the distance of electric field from the centre is${\mathrm{r}}_2=4 \mathrm{cm}\left(\frac{1 \mathrm{m}}{100\mathrm{cm}}\right)=4×{10}^{-2} \mathrm{m}$.
• For part (b) the distance of electric field from the centre is${\mathrm{r}}_3=9 \mathrm{cm}\frac{1 \mathrm{m}}{100\mathrm{cm}}=9×{10}^{-2} \mathrm{m}.$ .

The electric field is the area surrounded by the charged particles, and there is the force exerted in that area, which is either attracting or repelling in nature. The electric field is always generated due to the charged particles.

As the spheres are concentric and also has a radius of 2 cm , 5 cm and 10 cm then for the electric field from about 1 cm from the center of the sphere, the magnitude of the electric field of all the spheres is 0 as it lies inside the spheres.

Thus, the net electric field of the spheres at a distance of 1 cm is 0.

As the radius of the second and the third sphere is more than 4 cm which are 5 cm and 10 cm, then, the magnitude of the electric field of both the second and the third sphere which are$E_2$and$E_3$ is 0 as the point lies inside both of the spheres. But for the first sphere, the point lies outside so it has some value.

The expression for the magnitude of electric field for the first sphere is given as,

$E_1=k\frac{q_1}{{r_2}^2}$

Here,$E_1$ is the magnitude of the electric field for the first sphere,k is the electric field constant, $q_1$is the charge of the first sphere and$r_2$ is the distance of the electric field from the center.

Substitute$9×{10}^9 \mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{C}}^2\mathrm{for}\mathrm{k},4×{10}^{-2} \mathrm{m}\mathrm{for}{\mathrm{r}}_2,\mathrm{and}6×{10}^{-9} \mathrm{C}\mathrm{for}{\mathrm{q}}_1$in the above equation.

$$\begin{gathered} {\mathrm{E}}_1=(9×{10}^9 \mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{C}}^2) \frac{6×{10}^{-9} \mathrm{C}}{(4×{10}^{-2} \mathrm{m}{)}^2} \\ {\mathrm{E}}_1=33750 \mathrm{N}/\mathrm{C} \end{gathered}$$

The expression for net magnitude of electric field is given as,

$E_{net}=E_1+E_2+E_3$

Substitute 33750 N/C for $E_1$, and 0 for $E_2$, and 0 for $E_3$in the above equation.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{net}}=33750 \mathrm{N}/\mathrm{C}+0+0 \\ {\mathrm{E}}_{\mathrm{net}}=3.3×{10}^4 \mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the magnitude of electric field 4 cm at from the centre is $3.3×{10}^4 \mathrm{N}/\mathrm{C}$.

As the radius of the third sphere is more than 9 cm which is, 10 cm then, the magnitude of the electric field of the third sphere which is $E_3$is 0 as the point lies inside the sphere. But for the first and the second sphere, the point lies outside so it has some magnitude.

The expression for the magnitude of electric field for the first sphere is given as,

$E_1=k\frac{q_1}{{r_3}^2}$

Here, $E_1$is the magnitude of the electric field for the first sphere, k is the electric field constant, $q_1$is the charge of the first sphere and $r_3$is the distance of the electric field from the center.

Substitute $9×{10}^9 \mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{C}}^2\mathrm{for}\mathrm{k}\mathrm{k},9×{10}^{-2} \mathrm{m}\mathrm{for}{\mathrm{r}}_3,\mathrm{and}6×{10}^{-9} \mathrm{C}\mathrm{for}{\mathrm{q}}_1$in the above equation.

$$\begin{gathered} {\mathrm{E}}_1=(9×{10}^9 \mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{C}}^2) \frac{6×{10}^{-9} \mathrm{C}}{(9×{10}^{-2} \mathrm{m}{)}^2} \\ {\mathrm{E}}_1=6666.66 \mathrm{N}/\mathrm{C} \end{gathered}$$

The expression for the magnitude of electric field for the second sphere is given as,

$E_2=k\frac{q_2}{{r_3}^2}$

Here, $E_2$is the magnitude of the electric field for the second sphere, k is the electric field constant,$q_2$ is the charge of the second sphere and$r_3$ is the distance of the electric field from the center.

Substitute $9×{10}^9 \mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{C}}^2\mathrm{for}\mathrm{k},9×{10}^{-2} \mathrm{m}\mathrm{for}{\mathrm{r}}_3,\mathrm{and}-4×{10}^{-9} \mathrm{C}\mathrm{for}{\mathrm{q}}_2$in the above equation.

$$\begin{gathered} {\mathrm{E}}_2=(9×{10}^9 \mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{C}}^2) \frac{-4×{10}^{-9}\mathrm{C}}{(9×{10}^{-2} \mathrm{m}{)}^2} \\ {\mathrm{E}}_2=-4444.44 \mathrm{N}/\mathrm{C} \end{gathered}$$

Expression for net magnitude of electric field is given as,

$E_{net}=E_1+E_2+E_3$

Substitute $6666.66 \mathrm{N}/\mathrm{C}\mathrm{for}{\mathrm{E}}_1,\mathrm{and}-4444.44 \mathrm{N}/\mathrm{C}\mathrm{for}{\mathrm{E}}_2,\mathrm{and}0\mathrm{for}{\mathrm{E}}_3$in the above equation.

$$\begin{gathered} {\mathrm{E}}_{\mathrm{net}}=6666.66 \mathrm{N}/\mathrm{C}+(-4444.44 \mathrm{N}/\mathrm{C})+0 \\ {\mathrm{E}}_{\mathrm{net}}=2.22×{10}^3 \mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the magnitude of electric field at 9 cm from the centre is $2.22×{10}^3 \mathrm{N}/\mathrm{C}$.