Question

Location C is$\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{}\mathit{m}$ from a small sphere that has a charge of$\mathbf{4}\mathbf{}\mathit{n}\mathit{C}$uniformly distributed on its surface. Location D is$\mathbf{0}\mathbf{.}\mathbf{06}\mathbf{}\mathit{m}$from the sphere. What is the change in potential along a path from C to D?

Short answer

The change in potential along a path from C to D is $-1200\text{V}$.

Step-by-step solution

Given information

The magnitude of the charge is, $q=4\text{nC} \text{or} 4\times {10}^{-9}\text{C}$.

The distance ofC from the sphere is, $r_C=0.02\text{m}$.

The distance ofD from the sphere is, $r_D=0.06\text{m}$.

Concept of potential difference

The potential difference of a moving charge is calculated by subtracting the initial potential from the final potential of the charge.

If the value of the potential difference is negative, it means that the value of potential is decreasing as the charge is moving.

Determining the change in potential

The formula for the change in the potential fromC to D is given by,

$$\begin{gathered} \Delta V=V_D-V_C \\ \Delta V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r_D}-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r_C} \\ \Delta V=\frac{1}{4\pi {\epsilon }_0}q\left(\frac{1}{r_D}-\frac{1}{r_C}\right) \end{gathered}$$

Here, the constant term is $\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9\text{N}·{\text{m}}^2{\text{/C}}^2$.

Putting the values,

$$\begin{gathered} \Delta V=\left(9\times {10}^9\text{N}·{\text{m}}^2{\text{/C}}^2\right)\left(4\times {10}^{-9}\text{C}\right)\left(\frac{1}{0.06\text{m}}-\frac{1}{0.02\text{m}}\right) \\ \Delta V=36\times \left(\frac{1}{0.06}-\frac{1}{0.02}\right)\text{N}·\text{m/C}\times \frac{1\text{V}}{1\text{N}·\text{m/C}} \\ \Delta V=36\times \left(\frac{1-3}{0.06}\right)\text{V} \\ \Delta V=-1200\text{V} \end{gathered}$$

The negative sign indicates that the potential is decreasing.

Hence, the change in potential is $-1200\text{V}$.