91Ó°ÊÓ

1. Radius of the plastic disk,R =64.0 m
2. Uniform surface charge density,$\mathrm{σ}=7.73{\text{fC/m}}^{\text{2}}$
3. Three of the quadrants of the disk are removed.
4. The electric potential is V = 0 at infinity.
5. Distance of the point from the center,D = 0.259 m

Substituting the value of charge from the concept of the surface charge density in the formula of the electric potential due to a charge at a point, we can get the required value of the potential on the x-axis by using the given data.

Formulae:

The charge on the disk due to uniform charge density, $dq=\left(2\mathrm{π}rdr\right)\mathrm{σ}$ (i)

The electric potential at the point P due to the disk, $V=∫dV={∫}_0^R\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{dq}{\sqrt{r^2+D^2}}$ (ii)

So, the electric potential at P due to the complete disk using equation (i) in equation (ii) is given as follows:

$$\begin{gathered} V={∫}_0^R\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}\frac{\left(2\mathrm{Ï€}rdr\right)\mathrm{σ}}{\sqrt{r^2+D^2}} \\ =\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}{∫}_0^R\frac{rdr}{\sqrt{r^2+D^2}} \\ =\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}{\left[\sqrt{r^2+D^2}\right]}_0^R \\ ==\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\left[\sqrt{R^2+D^2}-D\right] \end{gathered}$$

So, the potential at P due to the ¼ slice of the disk is given by the above value as follows:

$$\begin{gathered} V_{net}=\frac{1}{4}V \\ =\frac{1}{8}\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\left[\sqrt{R^2+D^2}-D\right] \\ =\frac{1}{8}×\frac{7.73×{10}^{-15}}{8.85×{10}^{-12}}\left[\sqrt{{0.64}^2+{0.259}^2}-0.259\right] \\ =47.1×{10}^{-6}\text{V} \\ =47.1\mathrm{μ}\text{V} \end{gathered}$$

Hence, the value of the electric potential is $47.1\mathrm{μ}V$$47.1\mathrm{μ}V$.