Question

A dielectric lamina with charge density proportional to y covers the area between the parabola $\mathit{y}\mathbf{=}\mathbf{16}\mathbf{·}{\mathit{x}}^{\mathbf{2}}$ and the x axis. Find the total charge.

Short answer

The total charge of the dielectric lamina is,$546.13\mathrm{K}$.

Step-by-step solution

Definition of double integral and mass formula

The double integral of $\mathbf{f}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}$over the area Ain the (x,y) plane as the limit of this sum, and we write it as

Using the double integral the mass of the lamina:localid="1658895349906" $\mathbf{M}\mathbf{=}\mathbf{\int }{\mathbf{\int }}_{\mathbf{A}}\mathbf{\rho dA}$.

Diagram of the dielectric lamina

The figure of the dielectric lamina with charge density proportional to y covers the area between the parabola $y=16-x^2$ and x theaxis.

Calculation of the charge of the dielectric lamina

The total charge of the lamina is given by,

$$\begin{gathered} Q=\int \int A\mathrm{\rho dA} \\ ={\int }_{-4}^4\left[{\int }_0^{16-{\mathrm{x}}^2}\mathrm{Kydy}\right]\mathrm{dx} \\ =\mathrm{K}{\int }_{-4}^4\left({\overline{)\frac{1}{2}{\mathrm{y}}^2}}_0^{16-{\mathrm{x}}^2}\right)\mathrm{dx} \\ =\frac{\mathrm{K}}{2}{\int }_{-4}^4{\left(16-{\mathrm{x}}^2\right)}^2\mathrm{dx} \\ =\frac{\mathrm{K}}{2}{\int }_{-4}^4\left(256-32{\mathrm{x}}^2+{\mathrm{x}}^4\right)\mathrm{dx} \\ =\frac{\mathrm{K}}{2}{\overline{)\left[256\mathrm{x}-\frac{32}{3}{\mathrm{x}}^3+\frac{1}{5}{\mathrm{x}}^5\right]}}_{-4}^4 \\ =546.13\mathrm{K} \end{gathered}$$

where K is the proportionality constant.

Therefore, the total charge is,$546.13\mathrm{K}$.