91Ó°ÊÓ

The given data can be listed below as:

• The radius of the thin plastic spherical shell is,$\mathrm{r}=15\text{ c³¾}×\left(\frac{{\text{10}}^{\text{-2}}\text{″¾}}{\text{1 c³¾}}\right)=15×{\text{10}}^{\text{-2}}\text{″¾}$ .
• The charge of the thin plastic spherical shell is, $\mathrm{Q}=−8\text{ n°ä}(−8×{10}^{−9}\text{ C})$.
• The thickness of the block is, $\mathrm{l}=10\text{ c³¾}×\left(\frac{{\text{10}}^{\text{-2}}\text{″¾}}{\text{1 c³¾}}\right)=10×{\text{10}}^{\text{-2}}\text{″¾}$.
• The distance of the block from the sphere’s surface is,$s=10\text{ c³¾}×\left(\frac{{\text{10}}^{\text{-2}}\text{″¾}}{\text{1 c³¾}}\right)=10×{\text{10}}^{\text{-2}}\text{″¾}$ .

The electric field is a region that helps an electrically charged particle to exert force on another particle. The magnitude of the electric field is directly proportional to the charge of a particular object and inversely proportional to the distance of the object from the center of the electric field.

The diagram has been provided below:

As the charge on the sphere’s surface is uniform, then the charged sphere can be treated as a point charge having a negative sign. Hence, it attracts the positive charge to the block’s closer side and repels the negative charge to the block’s farther side.

The diagram has been provided below:

Because of the polarization of the mobile charges of the block by the charged sphere, then the negative charge gets accumulated at the right side and the positive charge gets accumulated at the left side of the block. As the electric field moves from the positive to the negative charge, then the electric field’s direction mainly points to the right side of the block.

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

$\mathrm{E}=\mathrm{k}\frac{\mathrm{Q}}{{(\mathrm{r}+\mathrm{l}/2)}^2}$

Here,$\mathrm{E}$is the magnitude of the electric field,$\mathrm{k}$is the electric field constant,$\mathrm{Q}$is the charge of the thin plastic spherical shell,$\mathrm{r}$is the radius of the thin plastic spherical shell and$l$is the thickness of the block.

Substitute the values in the above equation.

$\begin{array}{c}\mathrm{E}=(9×{10}^9\text{ N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})\frac{(−8×{10}^{−9}\text{ C})}{{((15×{10}^{−2}\text{″¾})+(10×{10}^{−2}\text{″¾})/2)}^2}\\ =\frac{(−72\text{ N}â‹\dots {\text{m}}^{\text{2}}\text{/C})}{{((15×{10}^{−2}\text{″¾})+(5×{10}^{−2}\text{″¾}))}^2}\\ =\frac{(−72\text{ N}â‹\dots {\text{m}}^{\text{2}}\text{/C})}{\left(0.04{\text{″¾}}^{\text{2}}\right)}\\ =−1800\text{ N/°ä}\end{array}$

The assumption made in this part is the electric field constant that is taken as$(9×{10}^9\text{ N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})$ .

Thus, the magnitude of the electric field vector is$−1800\text{ N/°ä}$ .

The assumption made in this part is the electric field constant that is taken as$(9×{10}^9\text{ N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})$ .