Question

Question: A plastic disk of radius R = 64.0 cmis charged on one side with a uniform surface charge density$\mathit{\sigma }\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{73}{\mathbf{\text{fC/m}}}^{\mathbf{\text{2}}}$, and then three quadrants of the disk are removed. The remaining quadrant is shown in Fig. 24-50.With V =0at infinity, what is the potential due to the remaining quadrant at point P, which is on the central axis of the original disk at distance D = 25.9 cmfrom the original center?

Short answer

Answer:

The potential at the point P, which is on the central axis of the original disk due to the remaining quadrant, is $47.1\mu V$.

Step-by-step solution

The given data

1. Radius of the plastic disk,R =64.0 m
2. Uniform surface charge density,$\sigma =7.73{\text{fC/m}}^{\text{2}}$
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

Understanding the concept of the electric potential

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, $dq=\left(2\pi rdr\right)\sigma$ (i)

The electric potential at the point P due to the disk, $V=\int dV={\int }_0^R\frac{1}{4\pi {\epsilon }_0}\frac{dq}{\sqrt{r^2+D^2}}$ (ii)

Calculation of the electric potential

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

$$\begin{gathered} V={\int }_0^R\frac{1}{4\pi {\epsilon }_0}\frac{\left(2\pi rdr\right)\sigma }{\sqrt{r^2+D^2}} \\ =\frac{\sigma }{2{\epsilon }_0}{\int }_0^R\frac{rdr}{\sqrt{r^2+D^2}} \\ =\frac{\sigma }{2{\epsilon }_0}{\left[\sqrt{r^2+D^2}\right]}_0^R \\ ==\frac{\sigma }{2{\epsilon }_0}\left[\sqrt{R^2+D^2}-D\right] \end{gathered}$$

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

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

Hence, the value of the electric potential is $47.1\mu V$$47.1\mu V$.