Question

A very thin spherical plastic shell of radius$\mathbf{15}\mathbf{\text{\hspace{0.17em}cm}}$ carries a uniformly distributed negative charge of $\mathbf{-}\mathbf{8}\mathbf{\text{\hspace{0.17em}nC}}\mathbf{(}\mathbf{-}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{\text{\hspace{0.17em}C}}\mathbf{)}$on its outer surface (so it makes an electric field as though all the charge were concentrated at the center of the sphere). An uncharged solid metal block is placed nearby. The block is$\mathbf{10}\mathbf{}\mathbf{cm}$ thick, and it is$\mathbf{10}\mathbf{}\mathbf{cm}$ away from the surface of the sphere. See Figure 14.97. (a) Sketch the approximate charge distribution of the neutral solid metal block.

(b) Draw the electric field vector at the center of the metal block that is due solely to the charge distribution you sketched (that is, excluding the contributions of the sphere).

(c) Calculate the magnitude of the electric field vector you drew. Explain briefly. If you must make any approximations, state what they are.

Short answer

a.

b.

c.

The magnitude of the electric field vector is$-1800\text{\hspace{0.17em}N/C}$ .

The assumption made in this part is the electric field constant that is taken as $(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})$.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The radius of the thin plastic spherical shell is,$\mathrm{r}=15\text{\hspace{0.17em}cm}\times \left(\frac{{\text{10}}^{\text{-2}}\text{\hspace{0.17em}m}}{\text{1\hspace{0.17em}cm}}\right)=15\times {\text{10}}^{\text{-2}}\text{\hspace{0.17em}m}$ .
• The charge of the thin plastic spherical shell is, $\mathrm{Q}=-8\text{\hspace{0.17em}nC}(-8\times {10}^{-9}\text{\hspace{0.17em}C})$.
• The thickness of the block is, $\mathrm{l}=10\text{\hspace{0.17em}cm}\times \left(\frac{{\text{10}}^{\text{-2}}\text{\hspace{0.17em}m}}{\text{1\hspace{0.17em}cm}}\right)=10\times {\text{10}}^{\text{-2}}\text{\hspace{0.17em}m}$.
• The distance of the block from the sphere’s surface is,$s=10\text{\hspace{0.17em}cm}\times \left(\frac{{\text{10}}^{\text{-2}}\text{\hspace{0.17em}m}}{\text{1\hspace{0.17em}cm}}\right)=10\times {\text{10}}^{\text{-2}}\text{\hspace{0.17em}m}$ .

Significance of the magnitude of the electric field

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.

 Step 3: (a) Sketching the appropriate charge distribution of the neutral solid metal block

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.

(b) Drawing the electric field vector at the center of the metal block

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.

(c) Determination of the magnitude of the electric field

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.

$\begin{array}{c}\mathrm{E}=(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})\frac{(-8\times {10}^{-9}\text{\hspace{0.17em}C})}{{((15\times {10}^{-2}\text{\hspace{0.17em}m})+(10\times {10}^{-2}\text{\hspace{0.17em}m})/2)}^2}\\ =\frac{(-72\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\text{/C})}{{((15\times {10}^{-2}\text{\hspace{0.17em}m})+(5\times {10}^{-2}\text{\hspace{0.17em}m}))}^2}\\ =\frac{(-72\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\text{/C})}{\left(0.04{\text{\hspace{0.17em}m}}^{\text{2}}\right)}\\ =-1800\text{\hspace{0.17em}N/C}\end{array}$

The assumption made in this part is the electric field constant that is taken as$(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})$ .

Thus, the magnitude of the electric field vector is$-1800\text{\hspace{0.17em}N/C}$ .

The assumption made in this part is the electric field constant that is taken as$(9\times {10}^9\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})$ .