Question

A uniform charge of $16.0\mu C$is on a thin circular ring lying in xy plane and centered on the origin. The ring’s radius is$3.00cm$. If point A is at the origin and point B is on the z-axis at $z=4.00cm$, what is$V_B-V_A$?

Short answer

The value of $V_B-V_A$is $-\text{1.92MV}$.

Step-by-step solution

Step 1: Given data:

The uniform charge,$q=+16.0\mu C$,

The radius of the circle,$R=0.03\text{m}$

The vertical distance, $AB=Z=\text{0.04 m}$.

Determining the concept:

Potential at a point due to a charge q is given as,

$\mathbf{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{r}}$

Here, is the distance of the point from the charge,$\mathbf{q}$ is the electric charge, and$\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}$is the Coulomb constanthaving a value $9\times 10N.m^2/C^2$.

Determining the VB-VA:

Due to the ring, the electric potential at$A$is,

$V_A=\frac{1}{4\pi {\epsilon }_o}·\frac{q}{R}$

$=\left(9.0\times {10}^{9}N.m^2/C^2\right)\times \frac{16.0\times {10}^{-6}C}{\left(0.03m\right)}$

$=4800\times {10}^3\text{V}$

$=4.\text{8 MV}$

Again, due to the ring, the electric potential at $B$is,

$V_B=\frac{1}{4\pi {\epsilon }_o}·\frac{q}{\sqrt{R^2+Z^2}}$
$V_B=$$\left(9.0\times {10}^{9}N.m^2/C^2\right)$$\times \frac{16\times {10}^{-6}\text{C}}{\sqrt{{0.03}^2+{0.04}^2}\text{m}}$

$=\left(9.0\times {10}^{9}N.m^2/C^2\right)$$\times \frac{16\times {10}^{-6}\text{C}}{\text{0.05 m}}$

$$\begin{gathered} =\text{2,880}\times {10}^3\text{V} \\ =2.8\text{8MV} \end{gathered}$$

Now,

$$\begin{gathered} V_B-V_A=\text{2.88MV}-\text{4.8MV} \\ =-\text{1.92MV} \end{gathered}$$

.

Hence, potential at $V_B-V_A$is $-\text{1.92MV}$.