Suppose that the radius of a disk is $\mathbf{R}\mathbf{=}\mathbf{20}$, and the total charge distributed uniformly all over the disk isrole="math" localid="1656058758873" $\mathbf{Q}\mathbf{=}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{C}$. Use the exact result to calculate the electric fieldfrom the center of the disk, and alsofrom the center of the disk. Does the field decrease significantly?

The given data can be listed below as-

The electric field describes as a particular region that is fruitful for the charged particle to exert force on another charged particle.

The equation of the electric field gives the electric field at the center of the disk.

The area of the disk can be expressed as,

Here, $\mathrm{R}$is the radius of the disk.

Forrole="math" localid="1656059658031" $\mathrm{R}=20\mathrm{cm}$ .

$A=\mathrm{\pi }{\left(0.2\mathrm{m}\right)}^2$

The equation of the electric field is expressed as,

$E=\frac{Q/A}{2{\epsilon }_0}\left[1-\frac{z}{{\left(R^2+Z^2\right)}^{{\displaystyle \frac{1}{2}}}}\right]$…(1)

Here, $Q$is the uniformly distributed charge on the disk, $A$is the area of the disk,$R$ is the disk’s radius and$Z$ is the distance of the electric field from the center of the ring.

For $Q=6\times {10}^{6}C$,$A=\mathrm{\pi }{\left(0.2\mathrm{m}\right)}^2$,${\epsilon }_0=8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2$,$\mathrm{Z}=1\mathrm{mm}$.

$$\begin{gathered} E=\frac{6\times {10}^6\mathrm{C}/\mathrm{\pi }{\left(0.2\mathrm{m}\right)}^2}{2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2}\left[1-\frac{1\mathrm{mm}\times {\displaystyle \frac{1\mathrm{m}}{1000\mathrm{mm}}}}{{\left({\left(0.2\mathrm{m}\right)}^2+{\left(1\mathrm{mm}\right)}^2\right)}^{{\displaystyle \frac{1}{2}}}}\right] \\ =\frac{6\times {10}^6\mathrm{C}/\mathrm{\pi }{\left(0.2\mathrm{m}\right)}^2}{2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2}\left[1-\frac{1\times {10}^{-3}\mathrm{m}}{{\left({\left(0.2\mathrm{m}\right)}^2+{\left(1\mathrm{mm}\times {\displaystyle \frac{1\mathrm{m}}{1000\mathrm{mm}}}\right)}^2\right)}^{{\displaystyle \frac{1}{2}}}}\right] \\ =\frac{6\times {10}^{6}C}{2.22\times {10}^{-12}{\mathrm{C}}^2/N}\left[1-\frac{1\times {10}^{-3}\mathrm{m}}{0.2\mathrm{m}}\right] \\ =2.702\times {10}^{18}\mathrm{N}/\mathrm{C}\times \left(1-5\times {10}^{-3}\right) \\ =2.702\times {10}^{18}\mathrm{N}/\mathrm{C}\times 0.995 \\ =2.68\times {10}^{18}\mathrm{N}/\mathrm{C} \end{gathered}$$

For$Q=6\times {10}^6\mathrm{C}$ ,$A=\mathrm{\pi }{\left(0.2\mathrm{m}\right)}^2$,${\epsilon }_0=8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2$,$\mathrm{z}=3\mathrm{mm}$.

$$\begin{gathered} E=\frac{6\times {10}^{6}C/\mathrm{\pi }{\left(0.2\mathrm{m}\right)}^2}{2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2}\left[1-\frac{1\mathrm{mm}\times {\displaystyle \frac{1\mathrm{m}}{1000\mathrm{mm}}}}{{\left({\left(0.2\right)}^2+{\left(1\mathrm{mm}\right)}^2\right)}^{{\displaystyle \frac{1}{2}}}}\right] \\ =\frac{6\times {10}^{6}C/\mathrm{\pi }{\left(0.2\mathrm{m}\right)}^2}{2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2}\left[1-\frac{1\times {10}^{-3}\mathrm{m}}{\left({\left(0.2\mathrm{m}\right)}^2+{\left(1\mathrm{mm}\times {\displaystyle \frac{1\mathrm{m}}{1000\mathrm{mm}}}\right)}^2\right){\displaystyle \frac{1}{2}}}\right] \\ =\frac{6\times {10}^{6}C/\mathrm{\pi }{\left(0.2\mathrm{m}\right)}^2}{2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/N}\left[1-\frac{3\times {10}^{-3}\mathrm{m}}{0.2\mathrm{m}}\right] \\ =2.702\times {10}^{18}\mathrm{N}/\mathrm{C}\times \left(1-0.985\right) \\ =2.66\times {10}^{18}\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the electric field at from the center of the disk is $2.68\times {10}^{18}\mathrm{N}/\mathrm{C}$and the electric field at from the center of the disk is.$2.66\times {10}^{18}\mathrm{N}/\mathrm{C}$. The field significantly decreases.