A circular disc of radius a has on one side a surface density of charge \(\sigma\) which varies with distance \(r\) from the centre but is constant for a given \(r\). What is the total charge on the disc of (a) \(\sigma=\sigma_{0} r / a\), (b) \(\sigma=\sigma_{0}\left(1-r^{2} / a^{2}\right)\), (c) \(\sigma=\sigma_{0} e^{-n i a}\) ?

To make the integration easier, we will convert the surface charge densities into polar coordinates. This means that, instead of integrating over x and y, we will integrate over r and θ (angle). The area element in polar coordinates is \(dA = r dr d\theta\).

For each case, we will integrate the surface charge density over the entire disc. In each case, the limits of integration for r will be from 0 to a and for θ will be from 0 to \(2\pi\). (a) \(\sigma_a = \sigma_0\frac{r}{a}\) Total charge on the disc, Q_a: \(Q_a = \int \int \sigma_a dA = \int_{0}^{a} \int_{0}^{2\pi} \left(\sigma_0 \frac{r}{a}\right) \cdot r dr d\theta\) (b) \(\sigma_b = \sigma_0\left(1-\frac{r^2}{a^2}\right)\) Total charge on the disc, Q_b: \(Q_b = \int \int \sigma_b dA = \int_{0}^{a} \int_{0}^{2\pi} \left(\sigma_0 \left(1-\frac{r^2}{a^2}\right)\right) \cdot r dr d\theta\) (c) \(\sigma_c = \sigma_0 e^{-ni\cdot a}\) Total charge on the disc, Q_c: \(Q_c = \int \int \sigma_c dA = \int_{0}^{a} \int_{0}^{2\pi} \left(\sigma_0 e^{-ni\cdot a }\right) \cdot r dr d\theta\)

Now, we will calculate the total charge for each case by solving the respective integrals. (a) Total charge, Q_a: \(Q_a = \int_{0}^{a} \int_{0}^{2\pi} \left(\sigma_0 \frac{r}{a}\right) \cdot r dr d\theta = \sigma_0\int_{0}^{a} \frac{r^2}{a} dr \int_{0}^{2\pi} d\theta = \frac{\sigma_0}{a}\left[\frac{r^3}{3}\right]_{0}^{a} \left[\theta\right]_{0}^{2\pi} = \frac{2\pi\sigma_0 a^2}{3}\) (b) Total charge, Q_b: \(Q_b = \int_{0}^{a} \int_{0}^{2\pi} \left(\sigma_0 \left(1-\frac{r^2}{a^2}\right)\right) \cdot r dr d\theta = \sigma_0 a^2\int_{0}^{a} \left(1-\frac{r^2}{a^2}\right) dr \int_{0}^{2\pi} d\theta = \sigma_0 a^2\left[\left[\frac{r^3}{3}\right]_{0}^{a} - \left[\frac{r^3}{3 a^2}\right]_{0}^{a}\right]\left[\theta\right]_{0}^{2\pi} = \frac{\pi\sigma_0 a^2}{2}\) (c) The given expression of \(\sigma\), \(\sigma_c = \sigma_0 e^{-ni\cdot a}\), is a complex function, which means that it doesn't represent a physical situation. Since it cannot represent a real surface charge density, it is not possible to obtain a total charge for this case.