A circular plastic disk with radius$\mathit{R}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.00}\mathbf{}\mathbf{\text{cm}}$has a uniformly distributed charge$\mathit{Q}\mathbf{=}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{.00}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{6}}\mathbf{)}\mathit{e}$on one face. A circular ring of widthis centered on that face, with the center of that width at radius$\mathit{r}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.50}\mathbf{}\mathbf{\text{cm}}$. In coulombs, what charge is contained within the width of the ring?

• Radius of the disk,$R=2.00\text{\hspace{0.17em}cm}$
• Charge distributed uniformly on one face of the disk,$Q=+\left(2.00x{10}^6\right)e$
• Width of the circular ring, centered on one face is$w=30\text{\hspace{0.17em}μm}$
• Radius of the ring,$r=0.50\text{cm}$

Using the concept of surface charge density, the relation on a surface with the density at that surface can be given. This determines the charge contained in the width of the ring.

Formula:

The surface charge density of a distribution, $\sigma =dq/dA$ (i)

We know that the area of the circular disk is given as:

$A=\pi r²$

$\Rightarrow dA=2\pi rdr$…… (a)

Now, the charge of an infinitesimal charge$dq$is given using this small area value and equation (i) as:

role="math" localid="1661915197346" $dq=\left(\frac{Q}{\pi R^2}\right)2\pi rdr$

Now, substituting the given values in the above equation for$dr$being the width of the ring, we get the charge in the width as:

$\begin{array}{c}dq=\left(\frac{2\times {10}^6\times 1.6\times {10}^{-19}\text{C}}{\pi {\left(0.02m\right)}^2}\right)2\pi \left(0.005\text{\hspace{0.17em}m}\right)(30\times {10}^{-6}\text{\hspace{0.17em}m})\\ =2.4x{10}^{-16}\text{\hspace{0.17em}C}\end{array}$

Hence, the value of the charge in the width is.$2.4x{10}^{-16}\text{\hspace{0.17em}C}$